What is the sum of the series 1 + (1 + x) + ( | vedclass

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

Question

(C) The $k$-th term of the series is $T_k = 1 + x + x^2 + \dots + x^{k-1} = \frac{1 - x^k}{1 - x}$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k =

Vedclass · Accepted Answer

$\frac{n(1 - x) - x(1 - x^n)}{(1 - x)^2}$

Vedclass · Answer

(C) The $k$-th term of the series is $T_k = 1 + x + x^2 + \dots + x^{k-1} = \frac{1 - x^k}{1 - x}$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{1 - x^k}{1 - x}$.$S_n = \frac{1}{1 - x} \left[ \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} x^k \right]$.$S_n = \frac{1}{1 - x} \left[ n - \frac{x(1 - x^n)}{1 - x} \right]$.$S_n = \frac{n(1 - x) - x(1 - x^n)}{(1 - x)^2}$.

What is the sum of the series $1 + (1 + x) + (1 + x + x^2) + (1 + x + x^2 + x^3) + \dots$ up to $n$ terms?

Similar Questions

The sum of $1^3 + 2^3 + 3^3 + 4^3 + \dots + 15^3$ is

$\frac{1^3 + 2^3 + 3^3 + 4^3 + \dots + 12^3}{1^2 + 2^2 + 3^2 + 4^2 + \dots + 12^2} = $

Find the sum to $n$ terms of the series whose $n^{th}$ term is given by $a_n = n^2 + 2^n$.

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

$1+\sin x+\sin ^2 x+\sin ^3 x+\ldots+\infty=4+2 \sqrt{3}$ and $0 < x < \pi, x \neq \frac{\pi}{2}$,then $x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module