Question

Prove that :-

$\quad \leadsto \quad \bf \Gamma ( n + 1 ) = n \Gamma ( n )$

Hint :-

$\quad \leadsto \quad \displaystyle \bf \int_{0}^{\infty} \bf x^{n-1} e^{-x} \: \: dx = \Gamma ( n )$ ​

Answer 1

It would be better not to involve integral in the solution and make the solution more complicated if there's a simple approach for this.

We know that:

$n! = \Gamma(n+1)$

Also, we know that:

$n! = n\cdot (n-1)!$

Therefore by using above two results, we get:

$\Gamma(n+1) = n \cdot (n-1)!$

$\implies \Gamma(n+1) = n \cdot \Gamma(n)$

And our required result is proved.

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