# Proof Challenge by Akshat

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: $$|x| \in [0,1)$$

Note by Akshat Joshi
1 year, 2 months ago

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For $$|x|\in [0,1)$$, 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,

$\left(x\frac{\textrm d}{\textrm dx}\right)^n\frac 1{1-x}=\sum_{k=0}k^nx^k=x+2^nx^2+3^nx^3+\ldots$

- 1 year, 1 month ago