Sum Questions

Sum questions and solutions are provided here in such a way that students can easily understand them and solve similar problems quickly. Students can start to solve the questions given below and verify their answers with the detailed explanations provided. This way of solving mathematical problems related to finding the sum will help students to improve their numerical skills.

What is a sum in mathematics?

In mathematics, a sum is a result obtained through the addition of two or more numbers that may or may not follow a pattern. Some of the basic formulas for finding the sum of a specific set of numbers are listed below:

• Sum of first n natural numbers = n(n + 1)/2
• Sum of first n even numbers = n(n + 1)
• Sum of first n odd numbers = n²
• Sum of squares of first n natural numbers = [n(n + 1)(2n + 1)]/6
• Sum of cubes of first n natural numbers = [n²(n + 1)²]/4
• Sum of first n terms of an AP is S_n = (n/2) [2a + (n – 1)d]

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Sum Questions and Answers

1. Find the sum of the first 15 natural numbers.

Solution:

We know that,

The sum of the first n natural numbers = n(n + 1)/2

When n = 15,

S = 15(15 + 1)/2

= (15 × 16)/2

= 120

Therefore, the sum of the first 15 natural numbers is 120.

2. Calculate the sum: 873 + 722 + 196

Solution:

Given: 873 + 722 + 196

The sum can be calculated as:

Therefore, 873 + 722 + 196 = 1791

3. What is the sum of all two-digit positive numbers?

Solution:

Two-digit positive numbers are: 10, 11, 12, …, 98, 99

This is an arithmetic progression with a = 10, d = 1 and l = 99

As we know, the sum of an AP with first term a and last term l is:

S = (n/2) (a + l); n is the number of terms.

= (90/2) (10 + 99)

= 45 × 109

= 4905

Thus, the sum of all two-digit positive numbers is 4905.

4. Estimate the sum:

Solution:

= 1² + 2² + 3² + …. + 10² + 11²

We know that the sum of squares of first n natural numbers is Σn² = [n(n+1)(2n+1)]/6

Substitute n = 11 in the above formula:

= [11(11 + 1) (2 × 11 + 1)]/6

= [11 × 12 × (22 + 1)]/ 6

= 11 × 2 × 23

= 506

Therefore,

5. Calculate the sum of all odd numbers between 1 and 51.

Solution:

Odd numbers between 1 and 51 are: 3, 5, 7, 9, 11, …., 47, 49

This is an AP with a = 3, d = 2.

Here, the nth term is 49.

a_n = 49

a + (n – 1)d = 49

3 + (n – 1)(2) = 49

(n – 1)2 = 46

n – 1 = 23

n = 24

S = (n/2)(a + a_n)

= (24/2) (3 + 49)

= 12 × 52

= 624

Hence, the sum of all odd numbers between 1 and 51 is 624.

6. Find the sum of all even numbers from 1 to 39.

Solution:

Even numbers from 1 to 39 are: 2, 4, 6, 8, …., 36, 38

We know that the sum of the first n even numbers is S = n(n + 1)

Also, the number of even numbers from 1 to 39 = 19

Sum of all even numbers from 1 to 39 = 19(19 + 1) = 19 × 20 = 380

7. Find the sum of all three-digit natural numbers divisible by 3.

Solution:

As we know,

The smallest three-digit natural number divisible by 3 = 102

The largest three-digit natural number divisible by 3 = 999

So, three-digit numbers that are divisible by 3:

102, 105, 108, ….., 996, 999

It is an AP with a = 102 and d = 3.

Here, a_n = 999

a + (n − 1)d = 999

102 + (n − 1)3 = 999

3(n – 1) = 999 – 102 = 897

n – 1 = 897/3 = 299

n = 300

Sum of all numbers

= (n/2) (a + an)

= (300/2) (102 + 999)

= 150 × 1101

= 165150

Therefore, the sum of all three-digit numbers divisible by 3 is 165150.

8. Find the sum of n terms of the series 1³ + 3³ + 5³ + 7³ + ….

Solution:

1³ + 3³ + 5³ + 7³ +….+ n³ = ∑_r=1ⁿ (2r – 1)³

= ∑_r=1ⁿ (8r³ – 12r² + 6r – 1)

= 8∑_r=1ⁿ r³ -12 ∑_r=1ⁿ r² + 6∑_r=1ⁿ r – ∑_r=1ⁿ 1

= 8[(n(n + 1))2/4] – 12[n(n + 1)(2n + 1)/6] + [6(n(n + 1)/2] – n

= 2n2(n + 1)2 – 2n(n + 1)(2n + 1) + 3n(n + 1) – n

= 2n(n + 1)[n(n + 1) – 2n – 1] + n[3n + 3 – 1]

= 2n(n + 1)(n² – n – 1) + n(3n + 2)

= 2n(n + 1)n(n – 1) – 2n² + 3n²

= 2n²(n² – 1) + n2

= n²(2n² – 1)

Thus, 1³ + 3³ + 5³ + 7³ +…. up to n terms = n²(2n² – 1).

9. What is the sum of cubes of the first 10 natural numbers?

Solution:

As we know,

Sum of cubes of the first n natural numbers = n²(n + 1)²/4

When n = 10,

S = (10)² (10 + 1)²/ 4

= (100 × 11²)/4

= 25 × 121

= 3025

Hence, the sum of cubes of the first 10 natural numbers is 3025.

10. Find the following sum.

27.076 + 0.55 + 0.004

Solution:

27.076 + 0.55 + 0.004

Therefore, 27.076 + 0.55 + 0.004 = 27.630

Practice Questions on Sum

1. Find the sum of the series 1 + 2 + 3 + 4 + …. + 60.
2. What is the sum of all three-digit numbers that are divisible by 5?
3. Find the sum of all two-digit odd numbers.
4. Calculate the sum: 9834.87 + 19.0056 + 0.00444
5. Estimate the sum: 1/(2² – 1) + 1/(4² – 1) + 1/(6² – 1) + …. + 1/(20² – 1)