Question

Let $S_n=\sum _{k=1}^{4n}(-1{)}^{\frac{k(k+1)}{2}}$. 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 $1332$

$\sum _{k=1}^{4n}{(-1)}^{\frac{k(k+1)}{2}}{k}^2=-1^2-2^2+3^2+4^2-5^2-6^2+7^2+8^2-.......+{\left(4n\right)}^2=(-1^2+3^2-5^2+7^2-....+{(4n-1)}^2)+(-2^2+4^2-6^2+8^2-....+{\left(4n\right)}^2)=\underset{ntermsinA.P.}{\underset{}{2(4+12+20+....(8n-4))}}+2\underset{ntermsinA.P.}{\underset{}{(6+14+22+....(8n-2))}}=n\left(8n\right)+n(8n+4).....(usesumofA.P.formula)=n(16n+4)$
.
Substituting $n=8$ gives the sum as $1056$. Hence, option A is possible.
Similarly, substituting $n=9$ gives the sum as $1332$. Hence, option D is also possible.