Question #674b8

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Answer 1 · 2017-01-07T16:14:55

See below.

Calling $S_{n} = n \sum k = 1 {(- 1)}^{n - k} k^{2}$ $S_n=sum_(k=1)^n(-1)^(n-k)k^2$ we have

$S_{1} = 1 = \frac{1 \cdot 2}{2}$ $S_1=1 = (1cdot 2)/2$
$S_{2} = 3 = \frac{2 \cdot 3}{2}$ $S_2=3=(2 cdot 3)/2$

Supposing that is true for $n$ $n$

$S_{n} = \frac{n (n + 1)}{2}$ $S_n=(n(n+1))/2$ and knowing that

$S_{n} - S_{n - 1} = n$ $S_n -S_(n-1) = n$ we have that

$S_{n + 1} = \frac{n (n + 1)}{2} + n + 1 = \frac{(n + 1) (n + 2)}{2}$ $S_(n+1) = (n(n+1))/2+n+1=((n+1)(n+2))/2$

So the summation formula is true.