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. 1332
• C. 1088
• D. 1120

Answer 1

Answer: (A), (B)

1332

$S_n={\sum _{k=1}^{4n}(-1)}^{\frac{k(k+1)}{2}}{k}^2$
$S_n=1^2{2}^2+3^2+4^2-5^2-6^2+........-{(4n-3)}^2-{(4n-2)}^2+{(4n-1)}^2+{\left(4n\right)}^2$
$S_n=\begin{array}{cc}& (3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+({11}^2-9^2)+({12}^2-{10}^2)+\\ & ......+{(4n-1)}^2-{(4n-3)}^2+{\left(4n\right)}^2-{(4n-2)}^2\end{array}$
$S_n=2(1+3)+2(4+2)+2(7+5)+2(8+6)+......+2(4n-1+4n-3)+2(4n+4n-2)$
$S_n=2[1+2+3+.....+4n]=\frac{2\cdot 4n(4n+1)}{2}$
From A, $4n(4n+1)=1056$
$4n^2+n=264$
$4n^2+n–264=0$
$n=8$
From B, $4n(4n+1)=1088$ (Not possible)
From C, $4n(4n+1)=1120$ (Not possible)
From D, $4n(4n+1)=1332$
$n=9$