Answer
$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$
Work Step by Step
From exercise $35$ with $a_{k}=k^{3}$
$\sum_{k=1}^{n}\left(k^{3}-(k-1)^{3}\right)=n^{3}$
Expanding $(k-1)^{3}$
$\sum_{k=1}^{n}\left(k^{3}-k^{3}+3 k^{2}-3 k+1\right)=n^{3}$
Cancelling and reorganizing
$3 \sum_{k=1}^{n} k^{2}=n^{3}+3 \sum_{k=1}^{n} k-\sum_{k=1}^{n} 1$
Substituting the values from exercise $37 b$ Thus
$3 \sum_{k=1}^{n} k^{2}=n^{3}+3 \frac{n(n+1)}{2}-n$
Putting together the terms under a common denominator
$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$
