Question
If $\sum_{k=1}^{n}\left(\sum_{m=1}^{k} m^{2}\right)=a n^{4}+b n^{3}+c n^{2}+d n+e$ then(a) $\mathrm{a}=\frac{1}{12}$(b) $e=0$(c) $c=\frac{5}{12}$(d) $\mathrm{d}=\frac{1}{6}$
Step 1
Step 1: We are given the sum $\sum_{k=1}^{n}\left(\sum_{m=1}^{k} m^{2}\right)$ and we know that the sum of squares of first $k$ natural numbers is given by $\frac{k(k+1)(2k+1)}{6}$. Show more…
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