The sum to 20 terms of the series 2^2-3^2+4^2 | vedclass

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

Question

(C) The given series is $2^2-3^2+4^2-5^2+6^2-\ldots$ up to $20$ terms.This can be written as the sum of two series: $S = (2^2+4^2+6^2+\ldots 	ext{ to

Vedclass · Accepted Answer

$1310$

Vedclass · Answer

(C) The given series is $2^2-3^2+4^2-5^2+6^2-\ldots$ up to $20$ terms.This can be written as the sum of two series: $S = (2^2+4^2+6^2+\ldots 	ext{ to } 10 	ext{ terms}) - (3^2+5^2+7^2+\ldots 	ext{ to } 10 	ext{ terms})$.$S = \sum_{n=1}^{10} (2n)^2 - \sum_{n=1}^{10} (2n+1)^2$.$S = \sum_{n=1}^{10} [ (2n)^2 - (2n+1)^2 ]$.Using the identity $a^2-b^2 = (a-b)(a+b)$:$S = \sum_{n=1}^{10} (2n - 2n - 1)(2n + 2n + 1) = \sum_{n=1}^{10} (-1)(4n+1)$.$S = -[ 4 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 1 ]$.$S = -[ 4 	imes \frac{10 	imes 11}{2} + 10 ]$.$S = -[ 220 + 10 ] = -230$.

The sum to $20$ terms of the series $2^2-3^2+4^2-5^2+6^2-\ldots$ is equal to $........$.

Similar Questions

If $S_n = 1^3 + 2^3 + \ldots + n^3$ and $T_n = 1 + 2 + \ldots + n$,then

The sum to $n$ terms of $(2n - 1) + 2(2n - 3) + 3(2n - 5) + \dots$ is

The value of $\sum\limits_{r = 16}^{30} {(r + 2)(r - 3)}$ is equal to

$1 + 3 + 7 + 15 + 31 + \dots$ to $n$ terms =

What is the $6^{th}$ term of the sequence $2, 1\frac{3}{4}, 1\frac{5}{9}, \dots$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module