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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$3003$

Vedclass · Answer

(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 $S_{77} =$

Similar Questions

The sum of the first $20$ terms of the sequence $0.7, 0.77, 0.777, \dots$ is

The value of $5^{1/2} \cdot 5^{1/4} \cdot 5^{1/8} \cdots \infty$ is .......

What is the arithmetic mean of the first $n$ terms of the sequence $1 \times 3 \times 5, 3 \times 5 \times 7, 5 \times 7 \times 9, \dots$?

If $\sum_{k=1}^n \left( \sum_{m=1}^k m^2 \right) = an^4 + bn^3 + cn^2 + dn + e$,then which of the following is true?

$\sum\limits_{m = 1}^n {{m^2}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module