Question

Find the sum of 5^2+7^2+9^2+.....+39^2

Answer 1

Answer:

Step-by-step explanation:The formula for this question is:

Sn=[n(n+1)(2n+1)]÷6.

Since sum starts from 5, findS39 then deduct the squares of 1,2,3 and 4 from the S39.The answer is :20540-1^2- 2^2- 3^2- 4^2

=20510...which is your required answer.

Answer 2

5² + 7² + 9²  + ... + 39² = 10650

Given:

• 5² + 7² + 9²  + ... + 39²

To Find:

• Sum

Solution:

• ∑n² = n(n + 1)(2n + 1)/6
• ∑n = n(n + 1)/2

Step 1:

Understand the series:

5² + 7² + 9²  + ... + 39²

5 = 2 (1) + 3

7 = 2(2) + 3

39 = 2(18) + 3

Hence   ∑(2n + 3)²   where n = 1 to 18

Step 2:

Using identity (a + b)² = a² + 2ab + b²

∑(2n + 3)²  = ∑(4n² + 12n + 9)

= 4∑n² + 12∑n + ∑9

Step 3:

Substitute ∑n² = n(n + 1)(2n + 1)/6 and ∑n = n(n + 1)/2

= 4n(n + 1)(2n + 1)/6 + 12 n(n + 1)/2 + 9n

Step 4:

Substitute n = 18

= 4(18)(19)(37)/6 + 12 (18)(19)/2 + 9(18)

= 8436 + 2052 + 162

= 10650

Hence, 5² + 7² + 9²  + ... + 39² = 10650