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

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

Question

(B) The $n$-th term of the series is $T_n = n^2(n + 1) = n^3 + n^2$.The sum of $n$ terms is $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (k^3 + k^2) = \sum_

Vedclass · Accepted Answer

$\frac{n(n + 1)(3n^2 + 7n + 2)}{12}$

Vedclass · Answer

(B) The $n$-th term of the series is $T_n = n^2(n + 1) = n^3 + n^2$.The sum of $n$ terms is $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (k^3 + k^2) = \sum_{k=1}^n k^3 + \sum_{k=1}^n k^2$.Using the standard summation formulas $\sum k^3 = [\frac{n(n+1)}{2}]^2$ and $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$:$S_n = [\frac{n(n+1)}{2}]^2 + \frac{n(n+1)(2n+1)}{6}$$S_n = \frac{n(n+1)}{2} [\frac{n(n+1)}{2} + \frac{2n+1}{3}]$$S_n = \frac{n(n+1)}{2} [\frac{3n^2 + 3n + 4n + 2}{6}]$$S_n = \frac{n(n+1)(3n^2 + 7n + 2)}{12}$.

The sum of the series $1^2 \cdot 2 + 2^2 \cdot 3 + 3^2 \cdot 4 + \dots$ to $n$ terms is:

Similar Questions

The value of $1^2 + 3^2 + 5^2 + ....... + 25^2$ is

$A$ person has two parents (father and mother),four grandparents,eight great-grandparents,and so on. Find the total number of ancestors the person has up to the $10^{th}$ generation.

Maximum value of the sum of the arithmetic progression $50, 48, 46, 44, \dots$ is:

The sum of the squares of the first $n$ natural numbers exceeds their sum by $330$. Then,$n = $

If the variance of the terms in an increasing $A.P.$,$b_{1}, b_{2}, b_{3}, \ldots, b_{11}$ is $90$,then the common difference of this $A.P.$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module