If sum_{k=1}^n left( sum_{m=1}^k m^2 right) = | vedclass

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Question

(A) We know that $\sum_{m=1}^k m^2 = \frac{k(k+1)(2k+1)}{6} = \frac{2k^3 + 3k^2 + k}{6}$.Then,$\sum_{k=1}^n \left( \sum_{m=1}^k m^2 \right) = \frac{1}

Vedclass · Accepted Answer

$a = \frac{1}{12}$

Vedclass · Answer

(A) We know that $\sum_{m=1}^k m^2 = \frac{k(k+1)(2k+1)}{6} = \frac{2k^3 + 3k^2 + k}{6}$.Then,$\sum_{k=1}^n \left( \sum_{m=1}^k m^2 \right) = \frac{1}{6} \sum_{k=1}^n (2k^3 + 3k^2 + k)$.$= \frac{1}{6} \left[ 2 \sum k^3 + 3 \sum k^2 + \sum k \right]$.$= \frac{1}{6} \left[ 2 \frac{n^2(n+1)^2}{4} + 3 \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right]$.$= \frac{1}{12} n^2(n^2+2n+1) + \frac{1}{12} (2n^3+3n^2+n) + \frac{1}{12} (n^2+n)$.$= \frac{1}{12} (n^4 + 2n^3 + n^2 + 2n^3 + 3n^2 + n + n^2 + n) = \frac{1}{12} n^4 + \frac{4}{12} n^3 + \frac{5}{12} n^2 + \frac{2}{12} n$.Comparing with $an^4 + bn^3 + cn^2 + dn + e$,we get $a = \frac{1}{12}$,$b = \frac{1}{3}$,$c = \frac{5}{12}$,$d = \frac{1}{6}$,and $e = 0$.Thus,$a = \frac{1}{12}$ is correct.

If $\sum_{k=1}^n \left( \sum_{m=1}^k m^2 \right) = an^4 + bn^3 + cn^2 + dn + e$,then which of the following is true?

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