Find the sum to n terms of the series 1^{2} + | vedclass

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(A) The $k$-th term of the series is $a_{k} = \sum_{i=1}^{k} i^{2} = \frac{k(k+1)(2k+1)}{6}$.Expanding this,we get $a_{k} = \frac{2k^{3} + 3k^{2} + k}

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$\frac{n(n+1)^{2}(n+2)}{12}$

Vedclass · Answer

(A) The $k$-th term of the series is $a_{k} = \sum_{i=1}^{k} i^{2} = \frac{k(k+1)(2k+1)}{6}$.Expanding this,we get $a_{k} = \frac{2k^{3} + 3k^{2} + k}{6} = \frac{1}{3}k^{3} + \frac{1}{2}k^{2} + \frac{1}{6}k$.The sum to $n$ terms is $S_{n} = \sum_{k=1}^{n} a_{k} = \sum_{k=1}^{n} (\frac{1}{3}k^{3} + \frac{1}{2}k^{2} + \frac{1}{6}k)$.Using the standard summation formulas $\sum k^{3} = \frac{n^{2}(n+1)^{2}}{4}$,$\sum k^{2} = \frac{n(n+1)(2n+1)}{6}$,and $\sum k = \frac{n(n+1)}{2}$:$S_{n} = \frac{1}{3} \cdot \frac{n^{2}(n+1)^{2}}{4} + \frac{1}{2} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{1}{6} \cdot \frac{n(n+1)}{2}$.Factor out $\frac{n(n+1)}{12}$:$S_{n} = \frac{n(n+1)}{12} [n(n+1) + (2n+1) + 1] = \frac{n(n+1)}{12} [n^{2} + n + 2n + 2] = \frac{n(n+1)}{12} [n(n+1) + 2(n+1)]$.$S_{n} = \frac{n(n+1)(n+1)(n+2)}{12} = \frac{n(n+1)^{2}(n+2)}{12}$.

Find the sum to $n$ terms of the series $1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + \ldots$

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