Find the sum to n terms of the series 3 times | vedclass

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

Question

(C) The $n^{	ext{th}}$ term $a_n$ of the series is given by the product of the $n^{	ext{th}}$ terms of the two sequences $(3, 6, 9, \dots)$ and $(8,

Vedclass · Accepted Answer

$3n(n+1)(n+3)$

Vedclass · Answer

(C) The $n^{	ext{th}}$ term $a_n$ of the series is given by the product of the $n^{	ext{th}}$ terms of the two sequences $(3, 6, 9, \dots)$ and $(8, 11, 14, \dots)$.$a_n = (3n) 	imes (3n + 5) = 9n^2 + 15n$.To find the sum $S_n = \sum_{k=1}^n a_k$,we calculate:$S_n = \sum_{k=1}^n (9k^2 + 15k) = 9 \sum_{k=1}^n k^2 + 15 \sum_{k=1}^n k$.Using the standard summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$:$S_n = 9 	imes \frac{n(n+1)(2n+1)}{6} + 15 	imes \frac{n(n+1)}{2}$.$S_n = \frac{3n(n+1)(2n+1)}{2} + \frac{15n(n+1)}{2}$.$S_n = \frac{3n(n+1)}{2} (2n + 1 + 5) = \frac{3n(n+1)}{2} (2n + 6)$.$S_n = 3n(n+1)(n+3)$.

Find the sum to $n$ terms of the series $3 \times 8 + 6 \times 11 + 9 \times 14 + \dots$

Similar Questions

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

$\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i {\sum\limits_{k = 1}^j 1 } } = \dots$

If $\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \dots + \infty = \frac{\pi^4}{90}$,then the value of $\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \dots + \infty$ is

The $9^{th}$ term of the sequence $27, 9, 5\frac{2}{5}, 3\frac{6}{7}, \dots$ is $.....$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module