Question

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

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

Answer 1

Answer: (A), (D)

The correct options are
A $1056$
D $1332$

$S_n=\sum _{k=1}^{4n}(-1{)}^{\frac{k(k+1)}{2}}{k}^2$

$=-1^2-2^2+3^2+4^2-5^2-6^2+7^2+8^2+\cdots +(4n{)}^2$

$=\left[\right(3^2-1^2)+(7^2-5^2)+\cdots n\text{terms}]+\left[\right(4^2-2^2)+(8^2-6^2)+\cdots n\text{terms}]$

$=8+24+40+\cdots n\text{terms}+12+28+44+\cdots n\text{terms}$

$=8(1+3+5+7+\cdots n\text{terms})+4(3+7+11+15+\cdots n\text{terms})$

$=8\times \frac{n}{2}[2+(n-1\left)2\right]+4\times \frac{n}{2}[6+(n-1\left)4\right]$

$=\left(8n^2\right)+(8n^2+4n)$

$=4n(4n+1)$

$S_n=4n(4n+1)=1056$ satisfies for $n=8$
$S_n=4n(4n+1)=1332$ satisfies for $n=9$