Question

If $S_n={\sum }_1^{4n}(-1{)}^{\frac{k(k+1)}{2}}{k}^2$. Then, $S_n$ can take value(s)

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

Answer 1

Answer: (A), (D)

1332

Convert it into differences and use sum of n terms of an AP,
ie., $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
Now, $S_n={\sum }_{k=1}^{4n}(-1{)}^{\frac{k(k+1)}{2}}.k^2$
$=-(1{)}^2-2^2+3^2+4^2-5^2-6^2+7^2+8^2+\dots \dots$
$=(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\dots \dots$
$=\underset{nterms}{\underset{}{2\left\{\right(4+12+20\dots \dots )}}+\underset{nterms}{\underset{}{(6+14+22+\dots \dots )\}}}$
$=2\left[\frac{n}{2}\right\{2\times 4+(n-1)8\}+\frac{n}{2}\{2\times 6+(n-1)8\left\}\right]$
$=2\left[n\right(4+4n-4)+n(6+4n-4\left)\right]$
$=2[4n^2+4n^2+2n]=4n(4n+1)$
Here, $1056=32\times 33,1088=32\times 34$,
$1120=32\times 35,1332=36\times 37$
1056 and 1332 are possible answers