The sum of (n - 1) terms of the series 1 + (1 | vedclass

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

Question

(C) Let $T_k$ be the $k^{th}$ term of the series. The $k^{th}$ term is the sum of the first $k$ odd numbers,which is $T_k = k^2$.The sum of the first

Vedclass · Accepted Answer

$\frac{(n - 1)n(2n - 1)}{6}$

Vedclass · Answer

(C) Let $T_k$ be the $k^{th}$ term of the series. The $k^{th}$ term is the sum of the first $k$ odd numbers,which is $T_k = k^2$.The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}$.We need the sum of $(n - 1)$ terms,so we replace $n$ with $(n - 1)$ in the formula for $S_n$:$S_{n-1} = \frac{(n - 1)((n - 1) + 1)(2(n - 1) + 1)}{6}$$S_{n-1} = \frac{(n - 1)(n)(2n - 2 + 1)}{6}$$S_{n-1} = \frac{(n - 1)n(2n - 1)}{6}$.

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

Similar Questions

Find the sum of the sequence $7, 77, 777, 7777, \ldots$ to $n$ terms.

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

If $\frac{1^3+2^3+3^3+\ldots \text{ upto } n \text{ terms}}{1 \cdot 3+2 \cdot 5+3 \cdot 7+\ldots \text{ upto } n \text{ terms}} = \frac{9}{5}$,then the value of $n$ is

The $n$th term of the series $1+3+7+13+21+\ldots$ is $9901$. The value of $n$ is

If $1^2 + 2^2 + 3^2 + \dots + 2009^2 = (2009)(335)(4019)$ and $(1)(2009) + 2(2008) + 3(2007) + \dots + 2009(1) = (2009)(335)(x)$,then $x$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module