If the sum of the series 1^2 + 2 cdot 2^2 + 3 | vedclass

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

Question

(C) For an odd $n$,the series consists of $n$ terms: $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + \dots + 2 \cdot (n-1)^2 + n^2$.This can be written as th

Vedclass · Accepted Answer

$\frac{n^2(n + 1)}{2}$

Vedclass · Answer

(C) For an odd $n$,the series consists of $n$ terms: $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + \dots + 2 \cdot (n-1)^2 + n^2$.This can be written as the sum of the first $(n-1)$ terms plus the $n$-th term.Since $(n-1)$ is even,we use the given formula for even terms: $S_{n-1} = \frac{(n-1)(n-1+1)^2}{2} = \frac{(n-1)n^2}{2}$.Now,adding the $n$-th term $(n^2)$: $S_n = \frac{(n-1)n^2}{2} + n^2$.Factoring out $n^2$: $S_n = n^2 \left( \frac{n-1}{2} + 1 \right) = n^2 \left( \frac{n-1+2}{2} \right) = \frac{n^2(n+1)}{2}$.

If the sum of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots + 2 \cdot (n-1)^2 + n^2$ (when $n$ is odd) is to be determined,given that for even $n$,the sum is $\frac{n(n+1)^2}{2}$,find the sum when $n$ is odd.

Similar Questions

What is the sum of the series $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$?

If $\alpha, \beta$ are natural numbers such that $100^{\alpha} - 199\beta = (100)(100) + (99)(101) + (98)(102) + \ldots + (1)(199)$,then the slope of the line passing through $(\alpha, \beta)$ and the origin is:

Find the sum of the series $1 + 3 + 7 + 15 + 31 + \dots$ up to $n$ terms.

For any integer $n \geq 1$,the sum $\sum_{k=1}^n k(k+2)$ is equal to

The sum of the series $1+3+5^2+7+9^2+\ldots$ up to $40$ terms is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module