Question

If $S_n=\sum _2^{4n}(-1{)}^{\frac{k(k+1)}{2}}{k}^2.$ Then, $S_n$ can take value $(s)$

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

Answer 1

Answer: (A), (D)

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