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

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

Question

(A) The $n^{th}$ term of the series is given by $T_n = n(2n + 1)^2$.Expanding this,we get $T_n = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.To find the sum $

Vedclass · Accepted Answer

$\frac{n}{6}(n + 1)(6n^2 + 14n + 7)$

Vedclass · Answer

(A) The $n^{th}$ term of the series is given by $T_n = n(2n + 1)^2$.Expanding this,we get $T_n = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.To find the sum $S_n$,we calculate $\sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (4k^3 + 4k^2 + k)$.Using the standard summation formulas $\sum k^3 = \frac{n^2(n+1)^2}{4}$,$\sum k^2 = \frac{n(n+1)(2n+1)}{6}$,and $\sum k = \frac{n(n+1)}{2}$:$S_n = 4 \cdot \frac{n^2(n+1)^2}{4} + 4 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$.$S_n = n^2(n+1)^2 + \frac{2n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$.Factoring out $\frac{n(n+1)}{6}$,we get:$S_n = \frac{n(n+1)}{6} [6n(n+1) + 4(2n+1) + 3]$.$S_n = \frac{n(n+1)}{6} [6n^2 + 6n + 8n + 4 + 3]$.$S_n = \frac{n(n+1)}{6} (6n^2 + 14n + 7)$.

The sum to $n$ terms of the series $1 \cdot 3^2 + 2 \cdot 5^2 + 3 \cdot 7^2 + \dots$ is

Similar Questions

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 $S_{77} =$

Find the $20^{\text{th}}$ term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$ up to $n$ terms.

The sequence of natural numbers is divided into groups as follows: $(1), (2, 3), (4, 5, 6), (7, 8, 9, 10), \dots$. Find the sum of the numbers in the $n^{th}$ group.

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 series $1 + 3x + 6x^2 + 10x^3 + \dots \infty$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module