Let S n = k = 1 4 n ( - 1 ) k ( k + 1 ) 2 ? k 2 . Then S n can take value(s) [2013]

Question

Let $S_{n} = \sum_{k = 1}^{4 n} {(- 1)}^{\frac{k (k + 1)}{2}} k^{2} .$ Then $S_{n}$ can take value(s) [2013]

Smart Achievers · Accepted Answer

(1, 4)

$S_{n} = - 1^{2} - 2^{2} + 3^{2} + 4^{2} - 5^{2} - 6^{2} + \dots$

$S_{n}$

$= \sum_{r = 1}^{n} {(4 r - 1)}^{2} + \sum_{r = 1}^{n} {(4 r)}^{2} - \sum_{r = 1}^{n} {(4 r - 3)}^{2} - \sum_{r = 1}^{n} {(4 r - 2)}^{2}$

$= [\sum_{r = 1}^{n} {(4 r - 1)}^{2} - {(4 r - 3)}^{2}] + 4 [\sum_{r = 1}^{n} {(2 r)}^{2} - {(2 r - 1)}^{2}]$

$= 8 \sum_{r = 1}^{n} (2 r - 1) + 4 \sum_{r = 1}^{n} (4 r - 1)$

$= 8 [2 \frac{n (n + 1)}{2} - n] + 4 [4 \frac{n (n + 1)}{2} - n]$

$= 8 n^{2} + 8 n^{2} + 4 n = 16 n^{2} + 4 n$

$For n = 8, 16 n^{2} + 4 n = 1056$

$and for n = 9, 16 n^{2} + 4 n = 1332$

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