8,110 views
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

2 votes
2 votes

$G(x)=\frac{1}{(1-x)^2} = (1-x)^{-2}$

Now we can expand $(1-x)^{-2}$ using negative binomial series theorem

$(1-x)^{-2}$

$= 1 +2x + \left ( \frac{1}{1*2}\ 2*3\  \right )x^2+ \left ( \frac{1}{1*2*3}\ 2*3*4\  \right )x^3+ \left ( \frac{1}{1*2*3*4}\ 2*3*4*5\  \right )x^4+...$

$= 1x^0+ 2x^1 +3x^2 + 4x^3+...$

We can clearly see that $g(i)$ is $i+1$

$\therefore$ Option $B.$ is correct answer.

0 votes
0 votes

B is the correct option. Let us put values

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 

0 votes
0 votes

Simple Trial and Error Approach,

 

Let x=0,

 

Taking LHS,

 $\frac{1}{(1-0)^2} => 1$

 

Taking RHS,

$\sum g(i)x^i => g(0)x^0 + g(1)x^1 + ...+ g(n)x^n => g(0) + 0 + ... + 0$

 

LHS=RHS,

1=g(0)

 

g(i) = i+1, only satisfies. Therefore, B.

 

Answer:

Related questions

60 votes
60 votes
9 answers
1
gatecse asked Sep 21, 2014
13,260 views
What is the minimum number of ordered pairs of non-negative numbers that should be chosen to ensure that there are two pairs $(a,b)$ and $(c,d)$ in the chosen set such th...
31 votes
31 votes
2 answers
2
go_editor asked Nov 27, 2016
11,342 views
Consider the following expression grammar. The semantic rules for expression evaluation are stated next to each grammar production.$$\begin{array}{l|l} E\rightarrow numbe...