The sum of the series 1 + 2 times 3 + 3 times | 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 given by $T_n = n(2n - 1) = 2n^2 - n$.
The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} (2k^2 - k) = 2

Vedclass · Accepted Answer

$946$

Vedclass · Answer

(B) The $n^{th}$ term of the series is given by $T_n = n(2n - 1) = 2n^2 - n$.
The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} (2k^2 - k) = 2 \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k$.
Using the standard summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$,we get:
$S_n = 2 \left[ \frac{n(n+1)(2n+1)}{6} ight] - \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{3} - \frac{n(n+1)}{2}$.
For $n = 11$:
$S_{11} = \frac{11(12)(23)}{3} - \frac{11(12)}{2}$.
$S_{11} = 11 	imes 4 	imes 23 - 11 	imes 6$.
$S_{11} = 1012 - 66 = 946$.

The sum of the series $1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \dots$ up to the $11^{th}$ term is

Similar Questions

If the $A.M.$ of two numbers is greater than $G.M.$ of the numbers by $2$ and the ratio of the numbers is $4:1$,then the numbers are

If $m$ is the $A.M.$ of two distinct real numbers $l$ and $n$ $(l, n > 1)$ and $G_1, G_2,$ and $G_3$ are three geometric means between $l$ and $n$,then $G_1^4 + 2G_2^4 + G_3^4$ equals:

The difference between an integer and its cube is divisible by:

$\sqrt {\underbrace {111........1}_{200\,digits} - \underbrace {222.......2}_{100\,digits}} $ equals :-

The sum of infinite terms of the geometric progression $\frac{\sqrt{2} + 1}{\sqrt{2} - 1}, \frac{1}{2 - \sqrt{2}}, \frac{1}{2}, \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module