Prove that : \[\displaystyle x + 2^nx^2 + 3^nx^3 + 4^nx^4 + ...... = x\frac{d}{dx}x\frac{d}{dx}......(\textrm{n times})\frac{1}{1-x} \]

Note: \(|x| \in [0,1)\)

Prove that : \[\displaystyle x + 2^nx^2 + 3^nx^3 + 4^nx^4 + ...... = x\frac{d}{dx}x\frac{d}{dx}......(\textrm{n times})\frac{1}{1-x} \]

Note: \(|x| \in [0,1)\)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestFor \(|x|\in [0,1)\), we have,

\[\frac 1{1-x}=1+x+x^2+\ldots=\sum_{k=0}^\infty x^k\]

Applying the operator \(x\frac{\textrm d}{\textrm dx}\), we have,

\[\frac 1{1-x}=1+x+x^2+\ldots=\sum_{k=0}^\infty x^k\implies x\frac{\textrm d}{\textrm dx}\frac 1{1-x}=\sum_{k=0}^\infty x\cdot kx^{k-1}=\sum_{k=0}k^1x^k\]

Repeating this argument \(n\) times, we have,

\[\left(x\frac{\textrm d}{\textrm dx}\right)^n\frac 1{1-x}=\sum_{k=0}k^nx^k=x+2^nx^2+3^nx^3+\ldots\] – Prasun Biswas · 2 months, 3 weeks ago

Log in to reply