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 + \sum_{k=1}^n

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 $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n k^3 + \sum_{k=1}^n k^2$. Using the standard summation formulas,$S_n = \left[ \frac{n(n + 1)}{2} \right]^2 + \frac{n(n + 1)(2n + 1)}{6}$. Factoring out $\frac{n(n + 1)}{2}$,we get $S_n = \frac{n(n + 1)}{2} \left[ \frac{n(n + 1)}{2} + \frac{2n + 1}{3} \right]$. Simplifying the expression inside the brackets: $\frac{3n^2 + 3n + 4n + 2}{6} = \frac{3n^2 + 7n + 2}{6}$. Thus,$S_n = \frac{n(n + 1)(3n^2 + 7n + 2)}{12}$.

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

Similar Questions

If $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \dots$ to $n$ terms $= n(n+1) f(n)$,then $f(2) =$

If $2.5+5.9+8.13+11.17+\ldots$ to $n$ terms $=an^3+bn^2+cn+d$,then $a-b+c-d=$

The sum of the infinite series $1+\frac{2}{3}+\frac{7}{3^{2}}+\frac{12}{3^{3}}+\frac{17}{3^{4}}+\frac{22}{3^{5}}+\ldots$ is equal to

The sum of an infinite $G.P.$ with common ratio $r$ can be found:

What is the arithmetic mean of the first $n$ terms of the sequence $1 \times 3 \times 5, 3 \times 5 \times 7, 5 \times 7 \times 9, \dots$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module