Question

Prove this identity between two infinite sums (with x ∈ R and n! stands for factorial): X∞ n=0 x n n! !2 = X∞ n=0 (2x) n n!

Answer 1

SOLUTION

TO PROVE

The identity

$\displaystyle \sf{ { \bigg(\sum\limits_{n = 0}^{ \infty} \: \frac{ {x}^{n} }{n!} \bigg)}^{2} = \sum\limits_{n = 0}^{ \infty} \frac{ {(2x)}^{n} }{n!} }$

CONCEPT TO BE IMPLEMENTED

We are aware of the expansion of $\sf{ {e}^{x} \: }$ :

$\displaystyle \sf{ {e}^{x} = 1 + \frac{ {x}^{} }{1! } + \frac{ {x}^{2} }{2! } +\frac{ {x}^{3} }{3! } + ... + \frac{ {x}^{n} }{n! } + ..... }$

$= \displaystyle \sf{ \sum\limits_{n = 0}^{ \infty} \frac{ {x}^{n} }{n!} }$

PROOF

LHS

$= \displaystyle \sf{ { \bigg(\sum\limits_{n = 0}^{ \infty} \: \frac{ {x}^{n} }{n!} \bigg)}^{2}}$

$= \displaystyle \sf{ \bigg( {1 + \frac{ {x}^{} }{1! } + \frac{ {x}^{2} }{2! } +\frac{ {x}^{3} }{3! } + ... + \frac{ {x}^{n} }{n! } + .....\bigg)}^{2} }$

$\sf{ = { ({e}^{x} )}^{2} \: }$

$= \displaystyle \sf{ {e}^{2x} }$

$\displaystyle \sf{ = 1 + \frac{ {2x}^{} }{1! } + \frac{ {(2x)}^{2} }{2! } +\frac{ {(2x)}^{3} }{3! } + ... + \frac{ {(2x)}^{n} }{n! } + ..... }$

$\displaystyle \sf{= \sum\limits_{n = 0}^{ \infty} \frac{ {(2x)}^{n} }{n!} }$

= RHS

Hence proved

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