Question

If $S_n= \displaystyle \sum_2^{4n} (-1) ^\frac{k(k+1)}{2}\, k^2.$ Then, $S_n$ can take value $(s)$

• A. 1056
• B. 1088
• C. 1120
• D. 1332

Answer 1

Answer: (A), (D)

Now,
$S_{n} =\displaystyle\sum_{k=1}^{4 n}(-1)^{\frac{k(k+1)}{2}} \cdot k^{2} $
$=-(1)^{2}-2^{2}+3^{2}+4^{2}-5^{2}-6^{2}+7^{2}+8^{2}+\ldots$
$=\left(3^{2}-1^{2}\right)+\left(4^{2}-2^{2}\right)+\left(7^{2}-5^{2}\right)+\left(8^{2}-6^{2}\right)+\ldots$
$=\underbrace{2\{(4+6+12+\ldots)}_{n \text { terms }}+\underbrace{(6+14+22+\ldots)\}}_{n \text { terms }} $
$=2\left[\frac{n}{2}\{2 \times 4+(n-1) 8\}+\frac{n}{2}\{2 \times 6+(n-1) 8\}\right]$
$=2[n(4+4 n-4)+n(6+4 n-4)] $
$=2\left[4 n^{2}+4 n^{2}+2 n\right]=4 n(4 n+1)$
Here, $1056 =32 \times 33,1088=32 \times 34,$
$1120 =32 \times 35,1332=36 \times 37$
$1056$ and $1332$ are possible answers.