Sum of First n Odd Natural Numbers

Example 1 : 

Find the sum of

1 + 3 + 5 + ........ to 40 terms

Solution :

Using 1 + 3 + 5 + ........ to n terms = n², 

1 + 2 + 3 + ........ + 40 terms = 40²

= 1600

Example 2 : 

Find the value of

1 + 3 + 5 + ........ + 55

Solution :

Using 1 + 3 + 5 + ........ + l = [(l + 1)/2]²,

1 + 2 + 3 + ........ + 55 = [(55 + 1)/2]²

= [56/2]²

= 28²

= 784

Example 3 : 

Find the value of

21 + 23 + 25 + ........ + 99

Solution :

21 + 23 + 25 + ........ + 99 :

= (1 + 2 + 3 + ........ + 99) - (1 + 2 + 3 + ........ + 19)

Using 1 + 3 + 5 + ........ + l = [(l + 1)/2]²,

= [(99 + 1)/2]² - [(19 + 1)/2]²

= [100/2]² - [20/2]²

= 50² - 10²

= 2500 - 100

= 2400

Example 4 : 

Find the sum of

2 + 6 + 10 + ........ to 50 terms

Solution :

2 + 6 + 10 + ........ to 50 terms : 

= 2(1 + 3 + 5 + ........ to 50 terms)

Using 1 + 3 + 5 + ........ to n terms = n², 

= 2(50²)

= 2(2500)

= 5000

Example 5 : 

Find the sum of

2 + 6 + 10 + ........ + 246

Solution :

2 + 6 + 10 + ........ + 256 : 

= 2(1 + 3 + 5 + ........ + 123)

Using 1 + 3 + 5 + ........ + l = [(l + 1)/2]²,

= [(123 + 1)/2]²

= [124/2]²

= 62²

= 3844

Example 6 : 

If 1 + 3 + 5 + ........ + k = 1444, then find k. 

Solution :

1 + 3 + 5 + ........ + k = 1444

Using 1 + 3 + 5 + ........ + l = [(l + 1)/2]²,

 [(k + 1)/2]² = 1444

 [(k + 1)/2]² = 38²

(k + 1)/2 = 38

Multiply each side by 2.

k + 1 = 76

Subtract 1 from each side. 

k = 75

Example 7 : 

If 1 + 3 + 5 + ........ to k terms = 676, then find k. 

Solution :

1 + 3 + 5 + ........ to k terms = 676

Using 1 + 3 + 5 + ........ to n terms = n², 

k² = 676

k² = 26²

k = 26

Example 8 :

Find the average of first 25 odd natural numbers. 

Solution :

Using 1 + 3 + 5 + ........ to n terms = n², to find the sum of first 25 odd natural numbers. 

1 + 2 + 3 + ........ to 25 terms = 25²

= 625

Average of first 25 odd natural numbers : 

= (Sum of first 25 odd natural numbers)/25

= 625/25

= 25

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.