Question

Prove for every positive integer n that 2! * 4! * 6! ... (2n)! ≥ [(n + 1)]^n.

Answer 1

Answer:Given below

Step-by-step explanation:

Using mathematical induction

For n=1

$2!=2^1$

True for n=1

Assume it is true for n=k

$2!\cdot 4!\cdot 6!\cdot 8!.......2k!\geq \left ( k+1\right )^{k}$

For n=k+1

$2!\cdot 4!\cdot 6!\cdot 8!.......2k!2\left ( k+1\right )!\geq \left ( k+1\right )^{k}\dot \left ( 2k+2\right )!$

because value of $2!\cdot 4!\cdot 6!\cdot 8!.......2k!=\left ( k+1\right )^{k}$

$\geq \left ( k+1\right )^{k}\dot \left ( 2k+2\right )!$

$\geq \left ( k+1\right )^{k}\left [ 2\left ( k+1\right )\right ]!$

$\geq \left ( k+1\right )^{k}\left ( 2k+\right )!\left ( 2k+2\right )$

$\geq \left ( k+1\right )^{k+1}\left ( 2k+\right )!$

Therefore $2!\cdot 4!\cdot 6!\cdot 8!.......2k!2\left ( k+1\right )! must be greater than \left ( k+1\right )^{k+1}$

Hence it is true for n=k+1

$2!\cdot 4!\cdot 6!\cdot 8!.......2k!2\left ( k+1\right )!\geq \left ( k+1\right )^{k+1}$

Hence it is true for n=k