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3.E: Number Patterns (Exercises)


Exercise (PageIndex{1}): Hexagonal numbers (cornered)

Consider the hexagonal numbers are the sequence (1,6,15, 28,45,66 cdots.) Predict the n th term. Explain your prediction.

Answer

(2n^2-n).

Exercise (PageIndex{2}): Finite sum

For each of the following, find the sum and explain your reasoning. Please do not use any formula.

  1. (1+3+5+7+9+cdots +197+199)
  2. (1+ displaystyle frac{1}{2}+ displaystyle frac{1}{4}+cdots + displaystyle frac{1}{2^{16}}+displaystyle frac{1}{2^{17}})
Answer
  1. (1+3+5+7+9+···+197+199)

Notice that (1,3,5,7 ,cdots) terms of a sequence. This is an Arithmetic Sequence because the difference remains the same between the terms throughout the entire sequence. Hence,( a = 1 & , d = 2).

Consider,

(S_ n = 1+3+5+7+9+···+197+199)

(S n = 199+197+195+193+191+···+3+1)

By adding we get,

((2S_n = 200+200+200+200+200+···+200+200)

(2S_n = 100(200))

(S_n = ((100)/2))(200))

(S_n = (50)(200))

(S_n = 10000)

Hence, the sum of the sequence is (10000.)

2.

Exercise (PageIndex{3}): Proof by induction

Consider the sequence ( 4,10,16, 22, 28,,dots), assume that the pattern continues.

  1. Show that the (n^{th}) term of this sequence can be expressed as (6n-2).
  2. Prove by using induction for all integers ( n geq 1, 4+10+16+dots+(6n-2)=n(3n+1))
Answer

1.

TermFirst difference
4
106
166
226

Notice that the first difference is constant. Hence the (n^{th}) term is a linear function.

Let (t_n = a_n + b.)

Then we need to find (a, b).

First Equation: Let (n = 1)

(t_1 = 4)

(4 = a(1) + b)

(4 = a + b)

Second Equation: Let (n = 2)

(t_2 = 10)

(10 = a(2) + b)

(10 = 2a + b)

To find (a), we use (10 = 2a + b) and (- 4 = a + b). Therfore, (6 = a.)

Now to find (b), we use (a = 6) and (4 = a + b),

(4 = (6) + b)

(4 - 6 = b)

(-2 = b).

Hence,(t_n = 6n - 2.)

2. Step 1: Base Step: Show that this statement is true for the smallest value

Check statement is true for n = 1.

L.H.S = 4

R.H.S = n(3n + 1)

= (1)(3(1) + 1)

= (1)(3 + 1)

= (1)(4)

= 4

Hence, the statement is true for n = 1.

Step 2: Induction Assumption:

We shall assume that the statement is true for n = k.

4 + 10 + 16 + . + (6k − 2) = k(3k + 1)

Step 3: Induction:

We shall show that the statement is true for n = k + 1.

4 + 10 + 16 + . + (6k − 2) + (6 (k + 1) − 2) = (k+1)(3(k + 1) + 1)

Consider, L.H.S = 4 + 10 + 16 + . + (6k − 2) + (6 (k + 1) − 2)

= k(3k + 1) + (6 ( k + 1) - 2)

= k (3k + 1) + (6k + 6 - 2)

= k (3k + 1) + (6k + 4)

= 3k 2 + k + 6k + 4

= 3k 2 + 7k + 4

= (k + 1)(3k + 4)

Hence, the statement is true for n = k + 1

Therefore, by induction the statement is true, ∀n ∈ N.

Exercise (PageIndex{4}): Proof by induction

Consider the sequence ( 3,11,19, 27, 35,dots), assume that the pattern continues.

  1. Show that the (n^{th}) term of this sequence can be expressed as (8n-5).
  2. Prove by using induction for all integers ( ngeq 1, 3+11+19 dots + (8n-5)=4n^2-n.)

Exercise (PageIndex{5}): Tribonacci

Let's start with the numbers (0,0,1,) and generate future numbers in our sequence by adding up the previous three numbers. Write out the first (15) terms in this sequence, starting with the first (1).

Exercise (PageIndex{6}): Proof by induction

The sequence (b_0,b_1,b_2....) is defined as follows: (b_0=1,b_1=3,b_2=5,) and for any integer (n geq 3, , b_n=3b_{n-2}+2b_{n-3}.)

  1. Find (b_3,b_4,b_5) and (b_6).
  2. Prove that (b_n < 2^{n+1}) for all integers (n geq 1.)

Exercise (PageIndex{7}): Quadratic Sequence

Find the (n^{th}) term of the sequence (5,10,17, 26, 37, cdots), assume that the pattern continues.

Answer

((n+1)^2+1=n^2+2n+2)

Exercise (PageIndex{8}): Proof by induction

Prove by using induction: for all integers ( ngeq 1, , 1+4+7 dots + (3n-2)=frac{n(3n-1)}{2}.)

Answer

Step 1: Base Step: Show that this statement is true for the smallest value

Check statement is true for n = 1.

L.H.S = 1

R.H.S = n(3n−1) / (2)

= (1)(3(1) − 1) / (2)

= (1)(3 − 1) / (2)

= (1)(2) / (2)

= 1

Hence, the statement is true for n = 1.

Step 2: Induction Assumption:

We shall assume that the statement is true for n = k.

1+4+7...+(3k−2)= k(3k−1) / (2)

Step 3: Induction:

We shall show that the statement is true for n = k + 1.

1+4+7...+ (3k - 2) + (3(k + 1) − 2) = (k + 1)(3(k + 1) − 1) / (2)

Consider, L.H.S = k(3k − 1) / (2) + (3 (k + 1) − 2)

= k (3k − 1) / (2) + (3k + 3) − 2)

= k (3k − 1) / (2) + (3k + 1)

= (3k 2 + k ) / (2) + (3k + 1)

= (3k 2 + k + 3k + 1) / (2)

= (3k 2 + 4k + 1) / (2)

= ((k + 1)(3k + 1)) / (2)

Hence, the statement is true for n = k + 1

Therefore, by induction the statement is true, ∀n ∈ N.

Exercise (PageIndex{9}): Recognizing sequence

Predict (n^{th}) term of the sequence (frac{2}{3},frac{3}{4}, frac{4}{5}cdots,) assume that the pattern continues. Explain your prediction.

Answer

( frac{n}{n+1}).

Exercise (PageIndex{10}): Recognizing sequence

Consider the sequence ( t_1=1, t_2=3+5, t_3=7+9+11, cdots ). Predict the n th term. Justify your prediction.

Exercise (PageIndex{11}): Proof by induction

Show that the perimeter of the design by joining (n) hexagons in a row is (8n+4) cm.

Exercise (PageIndex{13}): Pentagonal Numbers (cornered)

Find the (n^{th}) term of the sequence (1,5,12, 22, cdots),assume that the pattern continues.

Exercise (PageIndex{14}): Square Pyramidal numbers

Find the (n^{th}) term of the sequence (1,5,14,30 cdots)., assume that the pattern continues.

Exercise (PageIndex{15}): Difference

Compute the difference of each of the following sequences:

  1. (a_n= n^3)
  2. (a_n=n^{underline{3}})
  3. (a_n= displaystyle {n choose 3 } )
Answer
  1. (n^2+2n+1)
  2. (3 n^2 )
  3. (n choose 2).

Practicing Pattern exercises are always prescribed by many programmers as well as in books as it increases the ability to build logic while using Flow Control Statements. It also enhances logical thinking capabilities. In this article, We are going to see a list of Number patterns to practice for beginners and intermediate programmers.

Examples of Number Patterns in C language

Let us discuss some examples to understand the concept of number patterns in C easily.

Web development, programming languages, Software testing & others

Example #1

In the following C program, the user can enter a number of rows to print the number pyramid pattern as he wishes, then the result will be displayed on the screen:

#include<stdio.h>
#include<conio.h>
int main()
<
int n, i, j
printf("Enter the number of rows: ")
scanf("%d",&n)
for(i = 1 i <= n i++)
<
for(j = n j > i j--)
<
printf(" ")
>
for(j = 1 j <= i j++)
<
printf("%d ",j)
>
printf(" ")
>
return 0
>

Example #2

In the following C program, the user can enter the number of rows to print the half pyramid of numbers as he wishes, then the result will be displayed on the screen.

#include<stdio.h>
#include<conio.h>
int main()
<
int n, i, j
printf("Enter the number of rows: ")
scanf("%d",&n)
for(i = 1 i <= n i++)
<
for(j = 1 j <= i j++)
<
printf("%d",j)
>
printf(" ")
>
return 0
>

Example #3

In the following C program, the user can enter the number of rows to print the half pyramid of numbers as he wishes, then the result will be displayed on the screen.

#include<stdio.h>
#include<conio.h>
int main()
<
int n, i, j
printf("Enter the number of rows: ")
scanf("%d",&n)
for(i = 1 i <= n i++)
<
for(j = 1 j <= i j++)
<
printf("%d",i)
>
printf(" ")
>
return 0
>

Example #4

In the following C program, the user can enter the number of rows to print the Diamond pattern of numbers as he wishes, then the result will be displayed on the screen.

Example #5

In the following C program, the user can enter a number of rows to print the inverted half pyramid of numbers as he wishes, then the result will be displayed on the screen.

#include<stdio.h>
#include<conio.h>
int main()
<
int n, i, j
printf("Enter the number of rows: ")
scanf("%d",&n)
for(i = n i >= 1 i--)
<
for(j = 1 j <= i j++)
<
printf("%d",j)
>
printf(" ")
>
return 0
>

Example #6

In the following C program, the user can enter the number of rows to print the triangular number pattern as he wishes, then the result will be displayed on the screen:

Logic for the above program:

Between these two patterns spaces are printed in decreasing order. There are 10 spaces in 1 st row whereas 8 spaces in 2 nd row and so on the last row contains 0 spaces.

Example #7

In the following C program, the user can enter number of rows to print the number pyramid pattern as he wishes, then the result will be displayed on the screen:

#include<stdio.h>
#include<conio.h>
int main()
<
int i, s, n, j = 0, c = 0, c1 = 0
printf("Enter the number of rows: ")
scanf("%d",&n)
for(i = 1 i <= n ++i)
<
for(s = 1 s <= n-i ++s)
<
printf(" ")
++c
>
while(j != 2 * i - 1)
<
if (c <= n - 1)
<
printf("%d ", i + j)
++c
>
else
<
++c1
printf("%d ", (i + j - 2 * c1))
>
++j
>
c1 = c = j = 0
printf(" ")
>
return 0
>

Example #8

In the following C program, the user can enter number of rows to print the number pyramid pattern as he wishes, then the result will be displayed on the screen:

#include<stdio.h>
#include<conio.h>
int main()
<
int n, i, j, c = 1
printf("Enter the number of rows: ")
scanf("%d",&n)
for(i = 1 i <= n i++)
<
for(j = 1 j <= i ++j)
<
printf("%d ", c)
++c
>
printf(" ")
>
return 0
>

Example #9

In the following C program, the user can enter number of rows to print the Cross pattern of numbers as he wishes, then the result will be displayed on the screen:

#include<stdio.h>
#include<conio.h>
int main()
<
int n, i, j, c = 1
int m[5][5] = <0>
printf("Enter the number of rows: ")
scanf("%d",&n)
for(i = 1 i <= 5 i++)
<
for(j = 1 j <= 5 j++)
if(j == i || 6-i == j)
m[i-1][j-1] = c
if(i < 4) C
else --c
>
for(i = 0 i < 5 i++)
<
for(j = 0 j < 5 j++)
<
if(m[i][j] == 0)
printf(" ")
else
printf("%d",m[i][j])
>
printf(" ")
>
return 0
>

Example #10

In the following C program, the user can enter number of rows to print the Cross pattern of numbers as he wishes, then the result will be displayed on the screen:

#include<stdio.h>
#include<conio.h>
int main()
<
int n, i, j, c = 1
printf("Enter the number of rows: ")
scanf("%d",&n)
for(i = 1 i <= (2 * n) - 1 i++)
<
for (j = 1 j <= (2 * n) - 1 j++ )
<
if (i == j || i + j == 2 * n)
printf("%d", c)
else
printf(" ")
>
if (i < n)
C
else
c--
printf(" ")
>
return 0
>

Example #11

In the following C program, the user can enter number of rows to print the Square pattern of numbers as he wishes, then the result will be displayed on the screen:

#include<stdio.h>
#include<conio.h>
int main()
<
int n, i, j, c = 7, length = 18, max_length = 20
printf("Enter the number of rows: ")
scanf("%d",&n)
for(i = 1 i <= n i++)
<
for(j = 1 j <= n j++)
<
if(i == 1)
printf("% - 3d",j)
else if(j == n)
printf("% - 3d",C)
else if(i == n)
printf("% - 3d",length--)
else if(j == 1)
printf("% - 3d",max_length--)
else
printf(" ")
>
printf(" ")
>
return 0
>

Example #12

In the following C program, the user can enter number of rows to print the vertical triangle of numbers as he wishes, then the result will be displayed on the screen:

#include<stdio.h>
#include<conio.h>
int main()
<
int n, i, j
printf("Enter the number of rows: ")
scanf("%d",&n)
for(int i = 1 i < n i++)
<
for(int j = 1 j <= i j++)
printf("%d",j)
printf(" ")
>
for(int i = n i >= 0 i--)
<
for(int j = 1 j <= i j++)
printf("%d",j)
printf(" ")
>
return 0
>

Example #13

In the following C program, the user can enter a number of rows to print the vertical triangular of numbers as he wishes, then the result will be displayed on the screen:

#include<stdio.h>
#include<conio.h>
int main()
<
int n, i, j
printf("Enter the number of rows: ")
scanf("%d",&n)
for (int i = n i >= 0 i--)
<
for (int j = 1 j <= i j++)
printf("%d",j)
printf(" ")
>
for(int i = 1 i <= n i++)
<
for(int j = 1 j <= i j++)
printf("%d",j)
printf(" ")
>
return 0
>

Example #14

In the following C program, the user can enter the number of rows to print the Half Triangle pattern of numbers as he wishes, then the result will be displayed on the screen:

#include<stdio.h>
#include<conio.h>
int main()
<
int n, i, j, x, y
printf("Enter the number of rows: ")
scanf("%d",&n)
for (i = 1 i <= n i++)
<
if (i % 2 == 0)
<
x = 1
y = 0
>
else
<
x = 0
y = 1
>
for (j = 1 j <= i j++)
if (j % 2 == 0)
printf("%d",x)
else
printf("%d",y)
printf(" ")
>
return 0
>

Example #15

In the following C program, the user can enter the number of rows to print the inverted half pyramid pattern of numbers as he wishes, then the result will be displayed on the screen:

#include<stdio.h>
#include<conio.h>
int main()
<
int n, i, j
printf("Enter the number of rows: ")
scanf("%d",&n)
for(i = n i >= 1 i--)
<
for(j = i j >= 1 j--)
<
printf("%d", i)
>
printf(" ")
>
return 0
>

Recommended Articles

This is a guide to Number Patterns in C. Here we discuss the introduction and different examples along with the sample code. You can also go through our other suggested articles to learn more –


Number Patterns Grade 5 Worksheets

In the given series
20 , 19 , 17 , 14 , ________ , 5
20 – 1 = 19
19 – 2 = 17
17 – 3 = 14
We have observed that, every number is decreased by 1 , 2 , 3 and so on .
Therefore to get the missing number, we need to subtract 4 from the previous number
So, 14 – 4 = 10
Hence the missing number is 10
We can also see that subtracting next consecutive number i.e 5 from the missing number we get the next number of the series which is already given
10 – 5 = 5

Correct Answer – a) 10


Q.2) Explanation – Number Patterns Grade 5 Worksheets

In the given series
1600 , 1400 , 1200 , 1000 , 800 , ___ , ___ , ___
1600 – 200 = 1400
1400 – 200 = 1200
1200 – 200 = 1000
1000 – 200 = 800
We have observed that, every number is 200 less than the previous number. Therefore to get the next three numbers of the series we need to subtract 200 from the previous number
So, we can write the next numbers as
800 – 200 = 600
600 – 200 = 400
400 – 200 = 200
Therefore the next three numbers of the series are 600 , 400 , 200

Correct Answer – c) 600 , 400 , 200

Q.3) Explanation – Number Patterns Grade 5 Worksheets

In the given series
3 , 5 , 8 , 12 , ___ , 23
3 + 2 = 5
5 + 3 = 8
8 + 4 = 12
We have observed that, every number is increased by 2 , 3 , 4 and so on .
Therefore to get the missing number, we need to add 5 to the previous number 12
So, 12 + 5 = 17
Hence the missing number is 17
We can also see that adding next consecutive number i.e 6 to the missing number we get the next number of the series which is already given
17 + 6 = 23

Correct Answer – a) 17

Q.4) Explanation – Number Patterns Grade 5 Worksheets

In the given series
5 , 5 , 8 , 8 , 11 , __ , __
5 x 1 = 5
5 + 3 = 8
8 x 1 = 8
8 + 3 = 11
In the given series , the second number is 1 times the previous number and the number after the second number is 3 more than the previous number . Again the next number is 1 times the previous number and the number after it is 3 more than the previous number and so on .
Therefore to get the first missing number, we need to multiply the previous number by 1
So, 11 x 1 = 11
Now to get the next missing number, we need to add 3 to the previous number
11 + 3 = 14
Hence the missing numbers are 11 and 14

Correct Answer – a) 11, 14

Q.5) Explanation – Number Patterns Grade 5 Worksheets

In the given series
2 , 6 , 24 , 120 , __
2 x 3 = 6
6 x 4 = 24
24 x 5 = 120
We have observed that, every number is 3 times , 4 times , 5 times and so on to its previous number .
Therefore to get the missing number, we need to multiply 6 to the previous number .
So, 120 x 6 = 720
Hence the missing number is 720

Correct Answer – b) 720


Q.6) Explanation – Number Patterns Grade 5 Worksheets

In the given series
100 , 200 , 400 , 800 , ____ , ___
100 x 2 = 200
200 x 2 = 400
400 x 2 = 800
We have observed that, every number is 2 times the previous number. Therefore to get the next two numbers of the series we need to multiply the previous number by 2
So, we can write the next numbers as
800 x 2 = 1600
1600 x 2 = 3200
Therefore the next two numbers of the series are 1600 and 3200

Correct Answer – c) 1600 , 3200

Q.7) Explanation – Number Patterns Grade 5 Worksheets

In the given series
1 , 5 , 9 , 13 , 17 , __ , __ , __
1 + 4 = 5
5 + 4 = 9
9 + 4 = 13
13 + 4 = 17
We have observed that, every number is 4 more than the previous number. Therefore to get the next three numbers of the series we need to add 4 to the previous number
So, we can write the next numbers as
17 + 4 = 21
21 + 4 = 25
25 + 4 = 29
Therefore the next three numbers of the series are 21 , 25 , 29


3.2: Some Simple Patterns of Chemical Reactivity

Conceptual Problems

  1. What is the relationship between an empirical formula and a molecular formula
  2. Construct a flowchart showing how you would determine the empirical formula of a compound from its percent composition.

Numerical Problems

Please be sure you are familiar with the topics discussed in Essential Skills 2 ( Section 3.7 "Essential Skills 2" ) before proceeding to the Numerical Problems.

a. What is the formula mass of each species?

b. What is the molecular or formula mass of each compound?

1. What is the mass percentage of water in each hydrate?

2. What is the mass percentage of water in each hydrate?

3. Which of the following has the greatest mass percentage of oxygen&mdashKMnO4, K2Cr2O7, or Fe2O3?

4. Which of the following has the greatest mass percentage of oxygen&mdashThOCl2, MgCO3, or NO2Cl?

5. Calculate the percent composition of the element shown in bold in each compound.

6. Calculate the percent composition of the element shown in bold in each compound.

7. A sample of a chromium compound has a molar mass of 151.99 g/mol. Elemental analysis of the compound shows that it contains 68.43% chromium and 31.57% oxygen. What is the identity of the compound?

8. The percentages of iron and oxygen in the three most common binary compounds of iron and oxygen are given in the following table. Write the empirical formulas of these three compounds.

Compound % Iron % Oxygen Empirical Formula
1 69.9 30.1
2 77.7 22.3
3 72.4 27.6

9. What is the mass percentage of water in each hydrate?

10. What is the mass percentage of water in each hydrate?

11. Two hydrates were weighed, heated to drive off the waters of hydration, and then cooled. The residues were then reweighed. Based on the following results, what are the formulas of the hydrates?

Compound Initial Mass (g) Mass after Cooling (g)
NiSO4·xH2O 2.08 1.22
CoCl2·xH2O 1.62 0.88

12. Which contains the greatest mass percentage of sulfur&mdashFeS2, Na2S2O4, or Na2S?

13. Given equal masses of each, which contains the greatest mass percentage of sulfur&mdashNaHSO4 or K2SO4?

14. Calculate the mass percentage of oxygen in each polyatomic ion.

15. Calculate the mass percentage of oxygen in each polyatomic ion.

16. The empirical formula of garnet, a gemstone, is Fe3Al2Si3O12. An analysis of a sample of garnet gave a value of 13.8% for the mass percentage of silicon. Is this consistent with the empirical formula?

17. A compound has the empirical formula C2H4O, and its formula mass is 88 g. What is its molecular formula?

18. Mirex is an insecticide that contains 22.01% carbon and 77.99% chlorine. It has a molecular mass of 545.59 g. What is its empirical formula? What is its molecular formula?

19. How many moles of CO2 and H2O will be produced by combustion analysis of 0.010 mol of styrene?

20. How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0080 mol of aniline?

21. How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0074 mol of aspartame?

22. How many moles of CO2, H2O, N2, and SO2 will be produced by combustion analysis of 0.0060 mol of penicillin G?

23. Combustion of a 34.8 mg sample of benzaldehyde, which contains only carbon, hydrogen, and oxygen, produced 101 mg of CO2 and 17.7 mg of H2O.

a. What was the mass of carbon and hydrogen in the sample?

b. Assuming that the original sample contained only carbon, hydrogen, and oxygen, what was the mass of oxygen in the sample?

c. What was the mass percentage of oxygen in the sample?

d. What is the empirical formula of benzaldehyde?

e. The molar mass of benzaldehyde is 106.12 g/mol. What is its molecular formula?

24. Salicylic acid is used to make aspirin. It contains only carbon, oxygen, and hydrogen. Combustion of a 43.5 mg sample of this compound produced 97.1 mg of CO2 and 17.0 mg of H2O.

a. What is the mass of oxygen in the sample?

b. What is the mass percentage of oxygen in the sample?

c. What is the empirical formula of salicylic acid?

d. The molar mass of salicylic acid is 138.12 g/mol. What is its molecular formula?

25. Given equal masses of the following acids, which contains the greatest amount of hydrogen that can dissociate to form H + &mdashnitric acid, hydroiodic acid, hydrocyanic acid, or chloric acid?

26. Calculate the formula mass or the molecular mass of each compound.

a. heptanoic acid (a seven-carbon carboxylic acid)

b. 2-propanol (a three-carbon alcohol)

f. ethylbenzene (an eight-carbon aromatic hydrocarbon)

27. Calculate the formula mass or the molecular mass of each compound.

28. Given equal masses of butane, cyclobutane, and propene, which contains the greatest mass of carbon?

29. Given equal masses of urea [(NH2)2CO] and ammonium sulfate, which contains the most nitrogen for use as a fertilizer?

Conceptual Answers

1) What is the relationship between an empirical formula and a molecular formula

  • An empirical formula refers to the simplest ratio of elements that is obtained from a chemical formula while a molecular formula is calculated to show the actual formula of a molecular compound.

2) Construct a flowchart showing how you would determine the empirical formula of a compound from its percent composition.

Numerical Answers

a. What is the formula mass of each species?

b. What is the molecular or formula mass of each compound?

1. To two decimal places, the percentages are:

2. Percentage of Oxygen in each hydrates are:

4. % oxygen: ThOCl2, 5.02% MgCO3, 56.93% NO2Cl, 39.28%

5. To two decimal places, the percentages are:

9. To two decimal places, the percentages are:

10. What is the mass percentage of water in each hydrate?

14. Calculate the mass percentage of oxygen in each polyatomic ion.

16. The empirical formula of garnet, a gemstone, is Fe3Al2Si3O12. An analysis of a sample of garnet gave a value of 13.8% for the mass percentage of silicon. Is this consistent with the empirical formula?

No, the calculated mass percentage of silicon in garnet is 16.93%

19. How many moles of CO2 and H2O will be produced by combustion analysis of 0.010 mol of styrene?

20. How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0080 mol of aniline?

21. How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0074 mol of aspartame?

22. How many moles of CO2, H2O, N2, and SO2 will be produced by combustion analysis of 0.0060 mol of penicillin G?

24. Salicylic acid is used to make aspirin. It contains only carbon, oxygen, and hydrogen. Combustion of a 43.5 mg sample of this compound produced 97.1 mg of CO2 and 17.0 mg of H2O.

a. What is the mass of oxygen in the sample?

b. What is the mass percentage of oxygen in the sample?

c. What is the empirical formula of salicylic acid?

d. The molar mass of salicylic acid is 138.12 g/mol. What is its molecular formula?

26. Calculate the formula mass or the molecular mass of each compound.

27. To two decimal places, the values are:


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Analyse the diagram and complete the table.

Given a list of numbers: (7 4 1 -2 -5 ldots) determine the common difference for the list (if there is one).

egin d & = T_ <2>- T_ <1>= (4) - (7) = -3 & = T_ <3>- T_ <2>= (1) - (4) = -3 & = T_ <4>- T_ <3>= (-2) - (1) = -3 end All of the results are the same, which means we have found the common difference for these numbers: (d = -3).

For the pattern here: (- ext <0,55> ext <0,99> ext <2,49> ext <3,91> ldots) calculate the common difference.

If the pattern is not linear, write “no common difference”. Otherwise, give your answer as a decimal.

In this case the sequence is not linear. Therefore the final answer is that there is no common difference.

Consider the list shown here: (2 7 12 17 22 27 32 37 ldots)

If (T_ <5>= ext<22>) what is the value of (T_)?

Write down the next three terms in each of the following linear sequences:

(50r 46r 42r ldots)

egin d &= T_<2>-T_ <1> ext < or >T_<3>-T_<2> &= (46r) - (50r) ext < or >(42r) - (46r) &= -4r ext T_4 &= 38r T_5 &= 34r T_6 &= 30r end

Given a sequence which starts with the numbers: (6 11 16 21 ldots) determine the values of (T_ <6>) and ( T_<8>).

Given a list which starts with the letters: (A B C D ldots) determine the values of (T_<6>) and (T_<10>).

Find the sixth term in each of the following sequences:

(4 13 22 31 ldots)

egin d & = T_ <2>- T_ <1> & = 13 - 4 & = 9 end

Next we note that for each successive term we add (d) to the last term. We can express this as:

egin T_ <1>& = a = 4 T_ <2>& = a + d = 4 + 9 & = 4 + 9(1) T_ <3>& = T_ <2>+ d = 4 + 9 + 9 & = 4 + 9(2) T_ & = T_ + d = 4 + 9(n-1) & = 9n - 5 end

The general formula is (T_n = 9n - 5).

(5 2 -1 -4 ldots)

egin d & = T_ <2>- T_ <1> & = 2 - 5 & = -3 end

Next we note that for each successive term we add (d) to the last term. We can express this as:

egin T_ <1>& = a = 5 T_ <2>& = a + d = 5 + (-3) & = 5 + (-3)(1) T_ <3>& = T_ <2>+ d = 5 + (-3) + (-3) & = 5 + (-3)(2) T_ & = T_ + d = 5 + (-3)(n-1) & = 7 - 3n end

The general formula is (T_n = 7 - 3n).

Next we note that for each successive term we add (d) to the last term. We can express this as:

egin T_ <1>& = a = ext <7,4> T_ <2>& = a + d = ext <7,4>+ ext <2,3> & = ext <7,4>+ ext<2,3>(1) T_ <3>& = T_ <2>+ d = ext <7,4>+ ext <2,3>+ ext <2,3> & = ext <7,4>+ ext<2,3>(2) T_ & = T_ + d = ext <7,4>+ ext<2,3>(n-1) & = ext <7,4>+ ext<2,3>n - ext <2,3>= ext<2,3>n + ext <5,1>end

The general formula is (T_n = ext<2,3>n + ext<5,1>).

Find the general formula for the following sequences and then find T10, T15 and T30

Next we note that for each successive term we add (d) to the last term. We can express this as:

egin T_ <1>& = a = -18 T_ <2>& = a + d = -18 + (-4) & = -18 + (-4)(1) T_ <3>& = T_ <2>+ d = -18 + (-4) + (-4) & = -18 + (-4)(2) T_ & = T_ + d = -18 + (-4)(n-1) & = -4n - 14 end

The general formula is (T_n = -4n - 14).

Next we note that for each successive term we add (d) to the last term. We can express this as:

egin T_ <1>& = a = 1 T_ <2>& = a + d = 1 + (-7) & = 1 + (-7)(1) T_ <3>& = T_ <2>+ d = 1 + (-7) + (-7) & = 1 + (-7)(2) T_ & = T_ + d = 1 + (-7)(n-1) & = -7n + 8 end

The general formula is (T_n = -7n + 8).

The general term is given for each sequence below. Calculate the missing terms (each missing term is represented by (ldots)).

( ext <10> ldots ext <14> ldots ext <18>qquad T_n = 2 n + 8)

The missing terms are ( ext<12>) and ( ext<16>)

( ext <2> - ext <2> - ext <6> ldots - ext <14>qquad T_n = - 4n + 6)

egin T_n & = - 4 n + 6 T_4 & = - ext <4>( ext<4>) ext<+6> &= - ext <10>end

( ext <8> ldots ext <38> ldots ext <68>qquad T_n = 15 n - 7)

Find the general term in each of the following sequences:

(3 7 11 15 ldots)

egin d & = T_ <2>- T_ <1> & = 7 - 3 & = 4 end

Next we note that for each successive term we add (d) to the last term. We can express this as:

egin T_ <1>& = a = 3 T_ <2>& = a + d = 3 + 4 & = 3 + 4(1) T_ <3>& = T_ <2>+ d = 3 + 4 + 4 & = 3 + 4(2) T_ & = T_ + d = 3 + 4(n-1) & = 4n - 1 end

The general formula is (T_n = 4n - 1).

(-2 1 4 7 ldots)

egin d & = T_ <2>- T_ <1> & = 1 - (-2) & = 3 end

Next we note that for each successive term we add (d) to the last term. We can express this as:

egin T_ <1>& = a = -2 T_ <2>& = a + d = -2 + 3 & = -2 + 3(1) T_ <3>& = T_ <2>+ d = -2 + 3 + 3 & = -2 + 3(2) T_ & = T_ + d = -2 + 3(n-1) & = 3n - 5 end

The general formula is (T_n = 3n - 5).

(11 15 19 23 ldots)

egin d & = T_ <2>- T_ <1> & = 15 - 11 & = 4 end

Next we note that for each successive term we add (d) to the last term. We can express this as:

egin T_ <1>& = a = 11 T_ <2>& = a + d = 11 + 4 & = 11 + 4(1) T_ <3>& = T_ <2>+ d = 11 + 4 + 4 & = 11 + 4(2) T_ & = T_ + d = 11 + 4(n-1) & = 4n + 7 end

The general formula is (T_n = 4n + 7).

Next we note that for each successive term we add (d) to the last term. We can express this as:

The general formula is (T_n = frac<1><3>n).

Study the following sequence

Write down the next ( ext<3>) terms:

Find the general formula for the sequence.

Find the value of (n) if (T_n) is (- ext<917>).

What is the (346^< ext>) letter of the sequence:

The word “COMMON” is 6 letters long, so:

The remainder of 4 shows us that the (346^< ext>) letter is the (4th^< ext>) letter in the word, which is M

What is the (1000^< ext>) letter of the sequence:

The word “MATHEMATICS” is 11 letters long, so:

The remainder of 10 shows us that the (1000^< ext>) letter is the tenth letter in the word, which is C

The seating of a sports stadium is arranged so that the first row has ( ext<15>) seats, the second row has ( ext<19>) seats, the third row has ( ext<23>) seats and so on. Calculate how many seats are in the (25^< ext>) row.

We start by writing the given information as a sequence:

egin d & = T_ <2>- T_ <1> & = 19 - 15 & = 4 end

Next we note that for each successive term we add (d) to the last term. We can express this as:

egin T_ <1>& = a = 15 T_ <2>& = a + d = 15 + 4 & = 15 + 4(1) T_ <3>& = T_ <2>+ d = 15 + 4 + 4 & = 15 + 4(2) T_ & = T_ + d = 15 + 4(n-1) & = 4n + 11 end

The general formula is (T_n = 4n + 11).

The (25^< ext>) row is represented by (T_<25>). The number of seats in this row is:

egin T_ <25>& = 4(25) + 11 & = 111 end

There are 111 seats in the (25^< ext>) row.

The diagram below shows pictures which follow a pattern.

How many boxes will there be in the sixth picture?

Therefore three boxes are added each time and the sixth picture will have ( ext<17>) boxes

Determine the formula for the (n^< ext>) term.

The general term of the pattern is: (T_ = 3 n - 1).

Use the formula to find how many boxes are in the (30^< ext>) picture of the diagram.

egin T_ & = 3 n - 1 T_ <30>& = 3(30) - 1 longleftarrow ext < substitute n >= 30 & = 89 end

A single square is made from ( ext<4>) matchsticks. Two squares in a row need ( ext<7>) matchsticks and three squares need ( ext<10>) matchsticks.

Answer the following questions for this sequence.

We begin by writing a sequence to represent this:

We see from this that the first term is ( ext<4>).

Determine the common difference.

The common difference ((d)) is:

egin d & = T_ <2>- T_ <1> & = 7 - 4 & = 3 end

Determine the general formula.

To determine the general formula we note that for each successive term we add (d) to the last term. We can express this as:

egin T_ <1>& = a = 4 T_ <2>& = a + d = 4 + 3 & = 4 + 3(1) T_ <3>& = T_ <2>+ d = 4 + 3 + 3 & = 4 + 3(2) T_ & = T_ + d = 4 + 3(n-1) & = 3n + 1 end

The general formula is (T_n = 3n + 1).

A row has twenty-five squares. How many matchsticks are there in this row?

We note that a row with twenty-five squares is represented by (T_<25>). The number of matchsticks in this row is:

egin T_ <25>& = 3(25) + 1 & = 76 end

There are 76 matchsticks in the row with twenty-five squares.

You would like to start saving some money, but because you have never tried to save money before, you decide to start slowly. At the end of the first week you deposit ( ext, ext<5>) into your bank account. Then at the end of the second week you deposit ( ext, ext<10>) and at the end of the third week, ( ext, ext<15>). After how many weeks will you deposit ( ext, ext<50>) into your bank account?

We begin by writing down a sequence to represent this:

egin d & = T_ <2>- T_ <1> & = 10 - 5 & = 5 end

Now we note that for each successive term we add (d) to the last term. We can express this as:

egin T_ <1>& = a = 5 T_ <2>& = a + d = 5 + 5 & = 5 + 5(1) T_ <3>& = T_ <2>+ d = 5 + 5 + 5 & = 5 + 5(2) T_ & = T_ + d = 5 + 5(n-1) & = 5n end

The general formula is (T_n = 5n).

Now we need to find (n) such that (T_=50):

egin T_ & = 5n 50 & = 5n herefore n & = 10 end

After the tenth week you will deposit ( ext, ext<50>) into your bank account.

Consider the following list: [- 4y - 3 - y 2y + 3 5y + 6 8y + 9 ldots]

Find the common difference for the terms of the list. If the sequence is not linear (if it does not have a common difference), write “no common difference”.

egin d & = T_ <2>- T_ <1>= (- y) - (- 4 y - 3) = 3 y + 3 d & = T_ <3>- T_ <2>= (2 y + 3) - (- y) = 3 y + 3 d & = T_ <4>- T_ <3>= (5 y + 6) - (2 y + 3) = 3 y + 3 end The common difference for these numbers: (d = 3 y + 3).

If you are now told that (y = 1), determine the values of (T_<1>) and (T_<2>).

egin T_ <1>& = - 4 y - 3 & = - 4 (1)-3 & = -7 T_ <2>& = - y & = - (1) & = -1 end

If the following terms make a linear sequence: [2 n + frac<1> <2> 3 n + frac<5> <2> 7 n + frac<11> <2> ldots] Determine the value of (n). If the answer is a non-integer, write the answer as a simplified fraction.

egin T_ <2>- T_ <1>& = T_ <3>- T_ <2> left(3 n + frac<5><2> ight) - left(2 n + frac<1><2> ight) & = left(7 n + frac<11><2> ight) - left(3 n + frac<5><2> ight) 2left(3 n + frac<5><2> ight) - 2left(2 n + frac<1><2> ight) & = 2left(7 n + frac<11><2> ight) - 2left(3 n + frac<5><2> ight) 6 n + 5 - left(4 n + 1 ight) & = 14 n + 11 - left(6 n + 5 ight) 2 n + 4 & = 8 n + 6 -2 & = 6 n n & = - frac<1> <3>end

Now determine the numeric value of the first three terms. If the answers are not integers, write your answers as fractions.

egin ext T_ <1>& = 2 n + frac<1> <2> & = 2 left(- frac<1><3> ight)+frac<1> <2> & = - frac<1> <6> extT_ <2>& = 3 n + frac<5> <2> & = 3 left(- frac<1><3> ight)+frac<5> <2> & = frac<3> <2> extT_ <3>& = 7 n + frac<11> <2> & = 7 left(- frac<1><3> ight)+frac<11> <2> & = frac<19> <6>end The first three terms of this sequence are: (- frac<1> <6>, frac<3><2>) and (frac<19><6>).

How many blocks will there be in the (85^< ext>) picture?

Hint: Use the grey blocks to help.

The grey blocks can be represented by (n^2) and there are always ( ext<2>) white blocks.

Analyse the picture below:

How many blocks are there in the next picture?

Picture 4: (5^2 + 4 = 29) blocks

Write down the general formula for this pattern.

Picture 1: (2^2 + 1 qquad (n = 1))

How many blocks will there be in the 14th picture?

A horizontal line intersects a piece of string at ( ext<4>) points and divides it into five parts, as shown below.

If the piece of string is intersected in this way by ( ext<19>) parallel lines, each of which intersects it at ( ext<4>) points, determine the number of parts into which the string will be divided.

We need to determine a pattern for this scenario.

The first line divides the string into five parts. We can redraw the diagram to show the string with ( ext<2>) and ( ext<3>) lines:

From this we see that the two lines cut the string into ( ext<9>) pieces. Three lines cut the string into ( ext<13>) pieces. So for each line added we cut the line into ( ext<4>) more pieces.

So we can write the following sequence:

The common difference is ( ext<4>).

Next we note that for each successive term we add (d) to the last term. We can express this as:

egin T_ <1>& = a = 5 T_ <2>& = a + d = 5 + 4 & = 5 + 4(1) T_ <3>& = T_ <2>+ d = 5 + 4 + 4 & = 5 + 4(2) T_ & = T_ + d = 5 + 4(n-1) & = 4n + 1 end

The general formula is (T_n = 4n + 1).

When there are ( ext<19>) lines we are working with (T_<19>):

egin T_ <19>& = 4(19) + 1 & = 77 end

Therefore the string will be cut into ( ext<77>) parts.

Use a calculator to explore and then generalise your findings to determine the:

Therefore (3^<2007>) will follow the same pattern as the third row

therefore the units digit is ( ext<7>)

Therefore (7^<2008>) will follow the same pattern as the fourth row

therefore the tens digit is ( ext<0>)

remainder when (7^<250>) is divided by ( ext<5>)

Therefore (2^<250>) will follow the same pattern as the second row

therefore the remainder is ( ext<4>)

Analyse the diagram and complete the table.

The dots follow a triangular pattern and the formula is (T_n = frac<2>).

The general formula for the lines is (T_n = frac<3n(n - 1)><2>).

We are given the general formula for both the lines and the dots. We can determine the general formula for the sum of the lines and dots by adding the general formula for the lines to the general formula for the dots.


5 th GRADE MATH PRINTABLES

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    5 th Grade Integral Math Workbook


    Patterns with dots

    Some problems for the pattern can also involve a pattern of dots, where we need to find out the number and position of the dots in the pattern.

    For example:

    In the given examples, we found out the pattern by finding the dots that were added to the next figure.

    Number patterns are not restricted to a few types. They could be ascending, descending, multiples of a certain number, or series of even numbers, odd numbers etc.


    Pattern Worksheets

    These free worksheets will help your kids learn to recognize and complete patterns . They will also give practice recognizing the basic shapes and letters. Students discover the core (repeating part) of the pattern and then extend it in.

    Patterns of shapes
    Patterns of objects
    Patterns of letters

    Sample Kindergarten Patterns Worksheet

    K5 Learning offers free worksheets, flashcards and inexpensive workbooks for kids in kindergarten to grade 5. We help your children build good study habits and excel in school.


    Consider the sequence: 1, 4, 7, 10, … .What is the 10 th term in the sequence? What is the sum of the first ten terms?

    One day, Tom is very bored and he begins to record the hours his digital clock displays. He writes down the hour whenever an exact hour occurs (for example, he would write down number 17 when it is 5PM sharp). He starts at 8AM in the morning, and finishes at 3AM the next morning, when he goes to bed. What is the sum of all the numbers he writes down? Note that Tom’s clock shows 00:00 when midnight.


    Sequences Practice Questions

    1. A. Examine the sequence to find a pattern. The pattern is that every number is eight more than the last. Add to find the next number.

    2. B. A sequence of five consecutive even numbers is a sequence of even numbers such that the difference between one number and the next is always 2. Here are some examples:

    0, 2, 4, 6, 8
    12, 14, 16, 18, 20
    100, 102, 104, 106, 108

    One way to find the correct sequence is to set up and solve an equation. If x represents the first term, subsequent terms are x + 2, x + 4, x + 6, and x + 8. The sum of all the terms is 60, so

    Therefore, the sequence is 8, 10, 12, 14, 16.

    3. B. Examine the sequence to find a pattern. The pattern is that every number is half of the previous number. Multiply by to find the next number.

    4. C. As the problem states, the first two numbers are 0 and 1. Add them to find the third number in the Fibonacci sequence.

    Continue this process of adding consecutive terms in the sequence to find the next three numbers in the sequence.

    Therefore, the first six numbers in the Fibonacci sequence are 0, 1, 1, 2, 3, 5.

    5. E. As the problem states, the first number is 2. To find the second number, multiply 2 by 3.

    Continue this process of multiplying by 3 to find the next four numbers in the sequence.

    6 x 3 = 18
    18 x 3 = 54
    54 x 3 = 162
    162 x 3 = 486

    Therefore, the first six numbers in the sequence are 2, 6, 18, 54, 162, 486.

    6. A. A sequence of five consecutive even numbers is a sequence of even numbers such that the difference between one number and the next is always 2. Here are some examples:

    0, 2, 4, 6, 8
    12, 14, 16, 18, 20
    100, 102, 104, 106, 108

    If all of the numbers in a sequence are positive, then the sum of the sequence will also be positive. Therefore, in order for the sum to be zero, half of the numbers must be negative. In particular, for every positive number in the sequence is a corresponding negative number. Thus, the sequence should be -4, -2, 0, 2, 4.

    7. D. Examine the sequence to find a pattern. The pattern is that every number is six more than the last. Thus, a straightforward way to calculate the 23rd term is to write out the first 23 terms in the sequence, but this would be very tedious. A much easier method is to find a formula f(n) for the nth number in the sequence and then plug in 23 for n.

    Since each number is six more than the last, the formula will be something like f(n)=6n. However, notice that this formula is off by 3 for every number in the given sequence. Fix this by simply adding 3 in the formula. The result is f(n) = 6n + 3, which works for every value of n. Substitute 23 for n in this formula and calculate.

    8. C. Examine the sequence to find a pattern. The pattern is that every number is />times the previous number. Multiply 108 by />to find the next number in the sequence.

    Continue multiplying to find the next three numbers in the sequence.

    Therefore, the next four numbers are 162, 243, , .

    9. C. Examine the sequence to find a pattern. Calculate the difference between consecutive numbers.

    (2nd number) – (1st number) = 1 – 0 = 1
    (3rd number) – (2nd number) = 3 – 1 = 2
    (4th number) – (3rd number) = 6 – 3 = 3
    (5th number) – (4th number) = 10 – 6 = 4

    Therefore, the next number will be 5 more than the last number.

    10. B. Examine the sequence to find a pattern. The pattern is that every number is nine less than the previous number. Thus, a straightforward way to calculate the 31st term is to write out the first 31 terms in the sequence, but this would be very tedious. A much easier method is to find a formula f(n) for the nth number in the sequence and then plug in 31 for n.

    Since each number is nine less than the last, the formula will be something like f(n)=-9n. However, notice that this formula is off by 78 for every number in the given sequence. Fix this by simply adding 78 in the formula. The result is f(n) = -9n + 78, which works for every value ofn. To answer the question, substitute 31 for n in this formula and calculate.