Discussion Forum

Let   $S_{n}=\sum_2^{4n}(-1)^{\frac{k(k+1)}{2}}k^{2}$.Then , S_n can take value (s)

Answer:

Explanation:

Concept involved

It is to convert into differences and using sum of n terms of AP

 i.e,   $S_{n}=\frac{n}{2}[2a+(n-1)d]$

    $S_{n}=\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^{2}$

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

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

= 2{(4+6+12+...)  +(6+14+22+...)}

    n terms                    n terms

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

 =2[n(4+4n-4)+n(6+4n-4)]

=2[4n²+4n²+2n] =4n(n+1)

Here, 1056=32 x33, 1088=32 x34, 1120=32 x35, 1332=36x37

 1056  and 1332 are possible  answers