Hey just wondering if anyone could give me some help with this problem
prove by induction that:
$\displaystyle \binom{2n}{n}$≥ $\displaystyle \frac{4^n}{2\sqrt(n)}$
Hi pthomson75268,
If $\displaystyle \binom{2n}{n}\ \ge\ \frac{4^n}{2\sqrt{n}}$
then this statement is used to try to prove if $\displaystyle \binom{2(n+1)}{n+1}\ \ge\ \frac{4^{n+1}}{2\sqrt{n+1}}$
Proof
$\displaystyle \binom{2k+2}{k+1}=\frac{(2k+2)!}{[(2k+2)-(k+1)]!(k+1)!}=\frac{(2k+2)(2k+1)[(2k)!]}{(2k-k+1)!(k+1)!}$
$\displaystyle =\frac{(2k+1)(2k+1)[(2k)!]}{(2k-k+1)(2k-k)!(k+1)k!}=$$\displaystyle \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 $\displaystyle \ge\ \frac{4^{k+1}}{2\sqrt{k+1}}$
$\displaystyle \frac{4^{k+1}}{2\sqrt{k+1}}=\frac{4^k(4\sqrt{k})}{ 2\sqrt{k}\sqrt{k+1}}=\frac{4^k}{2\sqrt{k}}\ \left(\frac{4\sqrt{k}}{\sqrt{k+1}}\right)$
Therefore, if $\displaystyle \frac{(2k+2)(2k+1)}{(k+1)^2}\ \ge\ \frac{4\sqrt{k}}{\sqrt{k+1}}$
then $\displaystyle \binom{2(k+1)}{k+1}\ \ge\ \frac{4^{n+1}}{2\sqrt{k+1}}$ definately
Hence, the question now is... is $\displaystyle \frac{4k^2+6k+2}{(k+1)^2}\ \ge\ \frac{4\sqrt{k}\sqrt{k+1}}{k+1}$ ?
$\displaystyle \frac{4k^2+6k+2}{(k+1)^2}\ \ge\ \frac{4\sqrt{k}\sqrt{k+1}(k+1)}{(k+1)^2}$ ?
$\displaystyle 4k^2+6k+2\ \ge\ 4\sqrt{k}\sqrt{k+1}(k+1)$ ?
$\displaystyle \left(4k^2+6k+2\right)^2\ \ge\ 16k(k+1)(k+1)^2$ ?
$\displaystyle 16k^4+48k^3+52k^2+24k+4\ \ge\ 16k^4+48k^3+48k^2+16k$ ?
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.