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Let $G(x) = \frac{1}{(1-x)^2} = \sum\limits_{i=0}^\infty g(i)x^i$, where $|x| < 1$. What is $g(i)$?

1. $i$
2. $i+1$
3. $2i$
4. $2^i$
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Sumbody answer this ???b is the answer by putting x=0 but how to solve such types
+1

Why don't use the same method?

https://en.wikipedia.org/wiki/Taylor_series

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Yes but how to use compare them there are g(i) and Gx i know its easy but couldnt get it now
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$\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + x^5 + \dots + \infty$

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 $g(1) = 2, g(2) = 3 \color{red}{\Rightarrow g(i) = i+1}$

edited
+1
nice approach @mcjoshi sir
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Does someone has a list of all such important series. Knowing this series is essential to solve this question
We can use Maclaurin series $1/(1-x) = 1+x+x^{2}+x^{3}+x^{4}+...$

differentiating both side by x gives $1/(1-x)^{2} = 0+1+2x+3x^{2}+4x^{3}+...$

comparing this with given equaion $1/(1-x)^{2} = \sum g(i)x^{i} = g(0)+g(1)x+g(2)x^{2}+g(3)x^{3}+...$

$g(i)=i+1$

Option B

+1 vote
Using the extended Binomial theorem, we can write it as:

$(1-x)^{-2} = \sum_{i = 0}^{\infty}$${-n \choose k}(-x)^k$

From, this the coefficient can be written as: ${n+i-1} \choose i$, where $i = 2$

Therefore, $g(i) = \frac{(i+1)!}{i!} = i+1$
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How can you take combination of negative numbers, like you did in $-n \choose k$ and how did you arrive at $n+i-1 \choose i$? Please elaborate your answer.
+1

Check this out.

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