AP EAMCET · Maths · Sequences and Series
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 \(S_{77}=\)
- A − 3003
- B 3003
- C − 2926
- D 2926
Answer & Solution
Correct Answer
(B) 3003
Step-by-step Solution
Detailed explanation
\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\right) \\ & \quad-2\left(2^2+4^2+6^2+\ldots+76^2\right) \\ &…
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