Question

find the sum of the following series 6 square + 7 square + 8 square + ..... + 21 square​

Answer 1

Given:

A series: 6²+7²+8²+......+21²

To find:

Sum of the given series.

Solution:

We know that the sum of squares of n number is = [n(n+1)(2n+1)]/6

So first we would find the sum of the series:

1²+2²+3².......+21²

Here n = 21

The sum of this series will be:

S= [21 (21+1) (21*2 + 1)]/6 = [21*22*43]/ 6 = 3311

Now we would find the sum of the series:

1²+2²+3²+4²+5²

Here n =5

S' = [5(5+1)(2*5 + 1)]/6

= [5*6*11]/ 6 = 55

Now subtracting the two sums we would get the sum of the required series:

S- S' = 3311- 55 = 3256

Therefore the sum of the given series is 3256.

Answer 2

Answer:

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Step-by-step explanation:

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