The sum of the series 1 + 2x + 3x^2 + 4x^3 + | vedclass

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

Question

(A) Let $S_n$ be the sum of the given series to $n$ terms:$S_n = 1 + 2x + 3x^2 + 4x^3 + \dots + nx^{n - 1}$ $(i)$Multiply by $x$:$xS_n = x + 2x^2 + 3x

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $S_n$ be the sum of the given series to $n$ terms:$S_n = 1 + 2x + 3x^2 + 4x^3 + \dots + nx^{n - 1}$ $(i)$Multiply by $x$:$xS_n = x + 2x^2 + 3x^3 + \dots + (n - 1)x^{n - 1} + nx^n$ $(ii)$Subtracting $(ii)$ from $(i)$:$(1 - x)S_n = 1 + x + x^2 + x^3 + \dots + x^{n - 1} - nx^n$Using the sum formula for a geometric progression with $n$ terms:$(1 - x)S_n = \frac{1 - x^n}{1 - x} - nx^n$$(1 - x)S_n = \frac{1 - x^n - nx^n(1 - x)}{1 - x}$$(1 - x)S_n = \frac{1 - x^n - nx^n + nx^{n + 1}}{1 - x}$$(1 - x)S_n = \frac{1 - (n + 1)x^n + nx^{n + 1}}{1 - x}$Therefore,$S_n = \frac{1 - (n + 1)x^n + nx^{n + 1}}{(1 - x)^2}$.

The sum of the series $1 + 2x + 3x^2 + 4x^3 + \dots$ up to $n$ terms is:

Similar Questions

The $n^{th}$ term of the series $1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \dots$ will be

If $(10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + \dots + 10(11)^9 = k(10)^9$,then $k$ is equal to:

The sum of infinite terms of the following series $1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \dots$ will be

The sum $1 + 2 \cdot 3 + 3 \cdot 3^{2} + \dots + 10 \cdot 3^{9}$ is equal to

If $S(x) = (1+x) + 2(1+x)^2 + 3(1+x)^3 + \ldots + 60(1+x)^{60}$,$x \neq 0$,and $(60)^2 S(60) = a(b)^b + b$ where $a, b \in N$,then $(a+b)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module