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Let $G(x) = \large \frac{1}{(1 – x)^2} = \tt \sum_{i=0}^{\infty} g(i)x^i$, where $| x | < 1$. What is $g(i)$ ?

(A) i
(B) i+1
(C) 2i
(D) 2i
edited | 209 views

$\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + x^5 + \dots + \infty$

Differentiating it w.r.to $x$

$\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + \dots + \infty$

$\sum_{i=0}^{\infty} g(i)x^i = g(0) + g(1)x + g(2)x^2 + g(3)x^3 + \dots + \infty$

Comparing above two, we get $g(1) = 2, g(2) = 3 \color{red}{\Rightarrow g(i) = i+1}$
answered by Veteran (25k points)
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### S = 1 + 2x + 3x2 + 4x3 + .......... Sx =    x  + 2x2 + 3x3 + ..........  S - Sx = 1 + x + x2 + x3 + .... S - Sx = 1/(1 - x) [sum of infinite GP series with ratio < 1 is a/(1-r)] S = 1/(1 - x)2

answered by Loyal (3.2k points)

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