Question

If $S_n=\sum _{k=1}^{4n}{\left(-1\right)}^{\frac{k\left(k+1\right)}{2}}{k}^2$, then $S_n$ can take values

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

Answer 1

Answer: (A), (D)

$1332$

Explanation for the correct option

Step 1: Given information

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

Then,

$\sum _{k=1}^{4n}{\left(-1\right)}^{\frac{k\left(k+1\right)}{2}}{k}^2=-1^2-2^2+3^2+4^2-5^2-6^2+7^2+8^2.............+{\left(4n\right)}^2$

Step 2: Rearrange the numbers

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

$S_n=\left[-1^2+3^2-5^2+7^2................+{\left(4n-1\right)}^2\right]$

$S_n=2\left[4+12+20+........+\left(8n-4\right)\right]+2\left[6+14+22+....\left(8n-2\right)\right]$

Step 3: Now,

$2\left[4+12+20+........+\left(8n-4\right)\right]$ is an A.P of $n$ terms.

And,

$2\left[6+14+22+....\left(8n-2\right)\right]$$2\left[6+14+22+....\left(8n-2\right)\right]=n\left(8n+4\right)$ another A.P of $n$ terms.

Step 4: Sum of $n$ terms of the first and Second A.P is,

$2\left[4+12+20+........+\left(8n-4\right)\right]=n\left(8n\right)$

And,

$2\left[6+14+22+....\left(8n-2\right)\right]=n\left(8n+4\right)$

Step 5: Add both the A.P, we get,

$S_n=n\left(8n\right)+n\left(8n+4\right)$

$S_n=n\left(16n+4\right)$

Step 6: Substituting $n=8$ gives the sum as $1056$.

And, substituting $n=9$ gives the sum as $1332$.

Hence, option 'A' and option ‘D’ are correct.