Question

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

Answer 1

Answer:Given below

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