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27 votes
27 votes

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. $2^i$

8 Answers

Best answer
80 votes
80 votes

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

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

edited by
8 votes
8 votes
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
8 votes
8 votes
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, $g(i) = \frac{(i+1)!}{i!} = i+1$
Answer:

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