The sum,sum_{n=1}^{7} frac{n(n+1)(2n+1)}{4} i | vedclass

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(C) The given sum is $S = \sum_{n=1}^{7} \frac{n(n+1)(2n+1)}{4}$.We know that the sum of the first $n$ squares is $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2

Vedclass · Accepted Answer

$504$

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(C) The given sum is $S = \sum_{n=1}^{7} \frac{n(n+1)(2n+1)}{4}$.We know that the sum of the first $n$ squares is $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.Therefore,$n(n+1)(2n+1) = 6 \sum_{k=1}^{n} k^2$.Substituting this into the expression,we get $S = \frac{1}{4} \sum_{n=1}^{7} (6 \sum_{k=1}^{n} k^2) = \frac{6}{4} \sum_{n=1}^{7} \sum_{k=1}^{n} k^2$.Alternatively,we can expand the numerator: $n(n+1)(2n+1) = 2n^3 + 3n^2 + n$.So,$S = \frac{1}{4} \left( 2 \sum_{n=1}^{7} n^3 + 3 \sum_{n=1}^{7} n^2 + \sum_{n=1}^{7} n ight)$.Using the standard summation formulas:$\sum_{n=1}^{7} n^3 = \left( \frac{7 	imes 8}{2} ight)^2 = 28^2 = 784$.$\sum_{n=1}^{7} n^2 = \frac{7 	imes 8 	imes 15}{6} = 140$.$\sum_{n=1}^{7} n = \frac{7 	imes 8}{2} = 28$.Substituting these values: $S = \frac{1}{4} (2(784) + 3(140) + 28) = \frac{1}{4} (1568 + 420 + 28) = \frac{2016}{4} = 504$.

The sum,$\sum_{n=1}^{7} \frac{n(n+1)(2n+1)}{4}$ is equal to

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