# Math Help - proof by induction

1. ## proof by induction

Hey just wondering if anyone could give me some help with this problem

prove by induction that:

$\binom{2n}{n}$ $\frac{4^n}{2\sqrt(n)}$

2. Originally Posted by pthomson75268
Hey just wondering if anyone could give me some help with this problem

prove by induction that:

$\binom{2n}{n}$ $\frac{4^n}{2\sqrt(n)}$
Hi pthomson75268,

sorry, this was posted in error.
My work is below.

3. Originally Posted by pthomson75268
Hey just wondering if anyone could give me some help with this problem

prove by induction that:

$\binom{2n}{n}$ $\frac{4^n}{2\sqrt(n)}$
Hi pthomson75268,

If $\binom{2n}{n}\ \ge\ \frac{4^n}{2\sqrt{n}}$

then this statement is used to try to prove if $\binom{2(n+1)}{n+1}\ \ge\ \frac{4^{n+1}}{2\sqrt{n+1}}$

Proof

$\binom{2k+2}{k+1}=\frac{(2k+2)!}{[(2k+2)-(k+1)]!(k+1)!}=\frac{(2k+2)(2k+1)[(2k)!]}{(2k-k+1)!(k+1)!}$

$=\frac{(2k+1)(2k+1)[(2k)!]}{(2k-k+1)(2k-k)!(k+1)k!}=$ $\frac{(2k+2)(2k+1)}{k+1)(k+1)}\ \frac{(2k)!}{(2k-k)!k!}=\frac{(2k+2)(2k+1)}{(k+1)^2}\ \binom{2k}{k}$

We need to test if this is $\ge\ \frac{4^{k+1}}{2\sqrt{k+1}}$

$\frac{4^{k+1}}{2\sqrt{k+1}}=\frac{4^k(4\sqrt{k})}{ 2\sqrt{k}\sqrt{k+1}}=\frac{4^k}{2\sqrt{k}}\ \left(\frac{4\sqrt{k}}{\sqrt{k+1}}\right)$

Therefore, if $\frac{(2k+2)(2k+1)}{(k+1)^2}\ \ge\ \frac{4\sqrt{k}}{\sqrt{k+1}}$

then $\binom{2(k+1)}{k+1}\ \ge\ \frac{4^{n+1}}{2\sqrt{k+1}}$ definately

Hence, the question now is... is $\frac{4k^2+6k+2}{(k+1)^2}\ \ge\ \frac{4\sqrt{k}\sqrt{k+1}}{k+1}$ ?

$\frac{4k^2+6k+2}{(k+1)^2}\ \ge\ \frac{4\sqrt{k}\sqrt{k+1}(k+1)}{(k+1)^2}$ ?

$4k^2+6k+2\ \ge\ 4\sqrt{k}\sqrt{k+1}(k+1)$ ?

$\left(4k^2+6k+2\right)^2\ \ge\ 16k(k+1)(k+1)^2$ ?

$16k^4+48k^3+52k^2+24k+4\ \ge\ 16k^4+48k^3+48k^2+16k$ ?

Yes,

therefore, the trail of gunpowder has been put in place.
You only need light the match by proving the inequality for the first value of n.