What is the sum of the series 1^2 + 2.2^2 + 3 | vedclass

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

Question

(D) The given series is $S = 1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \dots + (2m-1)^2 + (2m)^2$. This can be split into two series: $S_1 = 1^2 + 3^2

Vedclass · Accepted Answer

$m(2m + 1)^2$

Vedclass · Answer

(D) The given series is $S = 1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \dots + (2m-1)^2 + (2m)^2$. This can be split into two series: $S_1 = 1^2 + 3^2 + 5^2 + \dots + (2m-1)^2$ $S_2 = 2^2 + 4^2 + 6^2 + \dots + (2m)^2 = 2^2(1^2 + 2^2 + 3^2 + \dots + m^2) = 4 	imes \frac{m(m+1)(2m+1)}{6} = \frac{2m(m+1)(2m+1)}{3}$. For $S_1$,the sum of squares of first $m$ odd numbers is $\frac{m(2m-1)(2m+1)}{3}$. Adding both: $S = \frac{m(2m+1)}{3} [ (2m-1) + 2(m+1) ] = \frac{m(2m+1)}{3} [ 2m - 1 + 2m + 2 ] = \frac{m(2m+1)(4m+1)}{3}$. Wait,re-evaluating the series structure: $1^2 + 2^2 + 3^2 + \dots + (2m)^2 = \frac{2m(2m+1)(4m+1)}{6} = \frac{m(2m+1)(4m+1)}{3}$.

What is the sum of the series $1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \dots + 2(2m)^2$?

Similar Questions

For all $n \in N$,the sum $S_n = 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}$ satisfies which of the following inequalities?

$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots$ to $16$ terms $=$

Find the sum of the series $1 + (1 + 2) + (1 + 2 + 3) + \dots$ up to $n$ terms.

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

Find the sum of the series $2 + 4 + 7 + 11 + 16 + \dots$ up to $n$ terms.

Vedclass Test Series

Exam Paper Generator

Online Exam Module