1^2+left(1^2+2^2right)+left(1^2+2^2+3^2right) | vedclass

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

Question

(C) The $n$-th term of the series is $T_n = \sum_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}{6}$.We need to find the sum $S_n = \sum_{i=1}^n T_i = \sum_{i=1}^n

Vedclass · Accepted Answer

$\frac{n(n+1)^2(n+2)}{12}$

Vedclass · Answer

(C) The $n$-th term of the series is $T_n = \sum_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}{6}$.We need to find the sum $S_n = \sum_{i=1}^n T_i = \sum_{i=1}^n \frac{i(i+1)(2i+1)}{6}$.Expanding the expression: $S_n = \frac{1}{6} \sum_{i=1}^n (2i^3 + 3i^2 + i) = \frac{1}{3} \sum i^3 + \frac{1}{2} \sum i^2 + \frac{1}{6} \sum i$.Using standard summation formulas:$S_n = \frac{1}{3} \left[\frac{n(n+1)}{2}\right]^2 + \frac{1}{2} \left[\frac{n(n+1)(2n+1)}{6}\right] + \frac{1}{6} \left[\frac{n(n+1)}{2}\right]$.Factoring out $\frac{n(n+1)}{12}$:$S_n = \frac{n(n+1)}{12} \left[ n(n+1) + (2n+1) + 1 \right] = \frac{n(n+1)}{12} [n^2 + n + 2n + 2] = \frac{n(n+1)(n^2+3n+2)}{12}$.Since $n^2+3n+2 = (n+1)(n+2)$,we get $S_n = \frac{n(n+1)^2(n+2)}{12}$.

$1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+\ldots+\left(1^2+2^2+\ldots+n^2\right)=$

Similar Questions

The sum of the series,$\frac{1}{2 \cdot 3} \cdot 2 + \frac{2}{3 \cdot 4} \cdot 2^{2} + \frac{3}{4 \cdot 5} \cdot 2^{3} + \ldots$ up to $n$ terms is

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 $S_{77} =$

What is the $99^{th}$ term of the series $2 + 7 + 14 + 23 + 34 + \dots$?

Consider an incomplete pyramid of balls on a square base having $18$ layers,and having $13$ balls on each side of the top layer. Then,the total number $N$ of balls in that pyramid satisfies

If the sum of $1 + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \dots$ to $n$ terms is $S$,then $S$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module