Question

Prove that $\frac{\left(2n\right)!}{2^{2n}(n!{)}^2}\le \frac{1}{\sqrt{3n+1}}$ for all $nϵN$.

Answer 1

P(n) : $\frac{\left(2n\right)!}{2^{2n}(n!{)}^2}\le \frac{1}{\sqrt{3n+1}}$

For n = 1

$\frac{2!}{2^2.1}\le \frac{1}{\sqrt{4}}$

$=\frac{1}{2}\le \frac{1}{2}$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

$\frac{\left(2n\right)!}{2^{2n}(n!{)}^2}\le \frac{1}{\sqrt{3n+1}}$ .....(1)

We have to show that,

$\frac{2(k+1)!}{2^{2(k+1)}\left[\right(k+1)!{]}^2}\le \frac{1}{\sqrt{3k+4}}Now,=\frac{2(k+1)!}{2^{2(k+1)}\left[\right(k+1)!{]}^2}=\frac{2(k+1)!}{2^{2k}{.2}^2(k+1)!(k+1)!}=\frac{(2k+2)(2k+1)\left(2k\right)!}{{4.2}^2(k+1)(k!)(k+1)(k!)}=\frac{2(k+1)(2k+1)\left(2k\right)!}{4.(k+1{)}^2{.2}^{2k}.(k!{)}^2}=\frac{2(2k+1)}{4(k+1)}.\frac{1}{\sqrt{3k}+1}\left[Usingequation\right(i\left)\right]=\frac{(2k+1)}{2(k+1)}.\frac{1}{\sqrt{3k}+1}=\frac{(2k+2)}{2(k+1)}.\frac{1}{\sqrt{3k+3+1}}=.\frac{1}{\sqrt{3k+4}}$

[Since 2k + 1 < 2k + 2 and 3k + 1 $\le$ 3k + 4]

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all $nϵN$ by PMI.