The sum of n terms of the following series 1 | 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 given by $T_k = 1 + x + x^2 + \dots + x^{k-1} = \frac{1 - x^k}{1 - x}$.The sum of $n$ terms is $S_n = \sum_{k=1}^

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

The sum of $n$ terms of the following series $1 + (1 + x) + (1 + x + x^2) + \dots$ will be

Similar Questions

If the roots of the equation ${x^3} - 12{x^2} + 39x - 28 = 0$ are in $A.P.$,then their common difference will be

The sum of the first $n$ natural numbers is:

The sums of $n$ terms of two arithmetic series are in the ratio $(2n + 3) : (6n + 5)$. Find the ratio of their $13^{th}$ terms.

$2.\overline{357} = $

The sum of all the products of the first $n$ natural numbers taken two at a time is

Vedclass Test Series

Exam Paper Generator

Online Exam Module