Let S_n = sum_{k=1}^n (-1)^{k-1} cdot k^2 for | vedclass

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(B) $S_n = 1^2 - 2^2 + 3^2 - 4^2 + \ldots + (-1)^{n-1} n^2$ \ $S_{77} = (1^2 - 2^2) + (3^2 - 4^2) + \ldots + (75^2 - 76^2) + 77^2$ \ Using the ident

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$3003$

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(B) $S_n = 1^2 - 2^2 + 3^2 - 4^2 + \ldots + (-1)^{n-1} n^2$ \ $S_{77} = (1^2 - 2^2) + (3^2 - 4^2) + \ldots + (75^2 - 76^2) + 77^2$ \ Using the identity $a^2 - b^2 = (a-b)(a+b)$,each pair $(k^2 - (k+1)^2) = (k - k - 1)(k + k + 1) = -(2k+1)$ \ $S_{77} = -(3 + 7 + 11 + \ldots + 151) + 77^2$ \ The sum inside the bracket is an arithmetic progression with $n = 38$ terms,first term $a = 3$,and last term $l = 151$ \ Sum $= \frac{38}{2}(3 + 151) = 19 	imes 154 = 2926$ \ $S_{77} = -2926 + 5929 = 3003$

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

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