Let S-n-0-n-n where -1- The value of - in the range 0-1- such that S-2- is

S=∑_n=0^∞nα S 0+1.α^1+2.α^2+3.α^3+...(i) α S=0+1α^2+2α^3+...(ii) ______________________________________ (1 α)S=0+α+α^2+α3+... (1 α)S=α/1 α ⇒ S=α/(1 α)^2 forS=2α ...

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Standard XII

Mathematics

Improper Integrals

Let S=∑n=0∞nα...

Question

Let $S = \sum_{n = 0}^{\infty} n α^{n}$ where $α$ <1. The value of $α$ in the range $0 < α < 1$ , such that S=2\alpha\) is

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Solution

$S = \sum_{n = 0}^{\infty} n α$
$S - 0 + 1. α^{1} + 2. α^{2} + 3. α^{3} + . . . (i)$
$α S = 0 + 1 α^{2} + 2 α^{3} + . . . (i i)$
______________________________________
$(1 - α) S = 0 + α + α^{2} + α 3 + . . .$
$(1 - α) S = \frac{α}{1 - α}$
$\Rightarrow S = \frac{α}{(1 - α)^{2}}$
for $S = 2 α = \frac{α}{(1 - α)^{2}}$
$\Rightarrow (1 - α)^{2} = \frac{1}{2}$
$\Rightarrow 1 - α = \frac{1}{\sqrt{2}}$
$\Rightarrow α = 1 - \frac{1}{\sqrt{2}} = 0.293$

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Let S=∑∞n=0nαn where α<1. The value of α in the range 0<α<1, such that S=2\alpha\) is

S=∑∞n=0nα S−0+1.α1+2.α2+3.α3+...(i) αS=0+1α2+2α3+...(ii) ______________________________________ (1−α)S=0+α+α2+α3+... (1−α)S=α1−α ⇒S=α(1−α)2 forS=2α=α(1−α)2 ⇒(1−α)2=12 ⇒1−α=1√2 ⇒α=1−1√2=0.293

Let $S = \sum_{n = 0}^{\infty} n α^{n}$ where $α$ <1. The value of $α$ in the range $0 < α < 1$ , such that S=2\alpha\) is