1.3k views

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. $2^i$
asked | 1.3k views
+1
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

0
Yes but how to use compare them there are g(i) and Gx i know its easy but couldnt get it now
0

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

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 Boss (28.4k points)
selected by
+1
nice approach @mcjoshi sir
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
answered by Active (2k points)

Option B) is the answer

answered by Loyal (8.1k points)
+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, $g(i) = \frac{(i+1)!}{i!} = i+1$
answered by Junior (849 points)
0
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.

### 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 (9.3k points)

1
2