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

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(C) We know that $\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$,$\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}$,and $\sum_{n=1}^{k} n^3 = \left(\frac{k(k+1)}{2}\

Vedclass · Accepted Answer

$504$

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(C) We know that $\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$,$\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}$,and $\sum_{n=1}^{k} n^3 = \left(\frac{k(k+1)}{2}ight)^2$. Given the sum is $\frac{1}{4} \sum_{n=1}^{7} (2n^3 + 3n^2 + n)$. Using the formulas for $k=7$: $\sum_{n=1}^{7} n = \frac{7 	imes 8}{2} = 28$. $\sum_{n=1}^{7} n^2 = \frac{7 	imes 8 	imes 15}{6} = 140$. $\sum_{n=1}^{7} n^3 = (28)^2 = 784$. Substituting these values: Sum $= \frac{1}{4} [2(784) + 3(140) + 28] = \frac{1}{4} [1568 + 420 + 28] = \frac{2016}{4} = 504$.

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

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