Let $G(x) = \frac{1}{(1-x)^2} = \sum\limits_{i=0}^\infty g(i)x^i$, where $|x| < 1$. What is $g(i)$?
Why don't use the same method?
https://en.wikipedia.org/wiki/Taylor_series
https://gateoverflow.in/88476/generating-function
$\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + x^5 + \dots + \infty$ 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 $g(1) = 2, g(2) = 3 \color{red}{\Rightarrow g(i) = i+1}$
Correct Answer: B
Option B) is the answer
@rawan Check this out.
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