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

If $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \ldots \text{ to } n \text{ terms} = n(n+1) f(n) - 3n$,then $f(1) =$

The sum of the series $3 + 33 + 333 + \dots$ to $n$ terms is

$2^2 + 4^2 + 6^2 + \dots + (2n)^2 = \dots$

Let $a, b, c, d$ be real numbers such that $\sum_{k=1}^n (a k^3+b k^2+c k+d)=n^4$,for every natural number $n$. Then,$|a|+|b|+|c|+|d|$ is equal to

The value of $1^3 - 2^3 + 3^3 - \dots + 15^3$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module