Let $S_n=\sum_{k=1}^n(-1)^{k-1} \cdot k^2$ for $n \geq 1$. Given that $S_{2 n}=-n(2 n+1)$ for $n=1,2,3, \ldots$. then $S_{77}=$

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Aicharya · Accepted Answer

$\begin{aligned} & S_n=\sum_{k=1}^n(-1)^{k-1} \cdot k^2 \ & S_n=1^2-2^2+3^2-4^2+\ldots(-1)^{n-1} n^2 \ & S_{77}=1^2-2^2+3^2-4^2+\ldots(-1)^{76} \cdot 77^2 \ & S_{77}=\left(1^2+2^2+3^2+\ldots+77^2ight) \ & \quad-2\left(2^2+4^2+6^2+\ldots+76^2ight) \ & =\left(1^2+2^2+3^2+\ldots+77^2ight) \ & \quad-2 \cdot 2^2\left(1^2+2^2+3^2+\ldots+38^2ight)\end{aligned}$ $\begin{aligned} & =\sum_{r=1}^{77} r^2-2 \cdot 2^2 \sum_{k=1}^{38} k^2 \ & =\frac{77 \cdot 78 	imes 155}{6}-\frac{8 \cdot 38 \cdot 39 \cdot 77}{6} \ & =\quad\left[	ext { as } \Sigma n^2=\frac{n(n+1)(2 n+1)}{6}ight] \ & =155155-152152=3003 .\end{aligned}$

Let $S_n=\sum_{k=1}^n(-1)^{k-1} \cdot k^2$ for $n \geq 1$. Given that $S_{2 n}=-n(2 n+1)$ for $n=1,2,3,…

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