# Solve the following

Question:

$\frac{(2 n) !}{2^{2 n}(n !)^{2}} \leq \frac{1}{\sqrt{3 n+1}}$ for all $n \in N$

Solution:

LetÂ P(n) be the given statement.

Thus, we have:

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

Step 1:

$P(1): \frac{2 !}{2^{2} .1}=\frac{1}{2} \leq \frac{1}{\sqrt{3+1}}$

Thus, $P(1)$ is true.

Step 2 :

Let $P(m)$ be true.

Thus, we have :

$\frac{(2 m) !}{2^{2 m}(m !)^{2}} \leq \frac{1}{\sqrt{3 m+1}}$

We need to prove that $P(m+1)$ is true.

Now,

$P(m+1):$

$\frac{(2 m+2) !}{2^{2 m+2}((m+1) !)^{2}}=\frac{(2 m+2)(2 m+1)(2 m) !}{2^{2 m} \cdot 2^{2}(m+1)^{2}(m !)^{2}}$

$\Rightarrow \frac{(2 m+2) !}{2^{2 m+2}((m+1) !)^{2}} \leq \frac{(2 m) !}{2^{2 m}(m !)^{2}} \times \frac{(2 m+2)(2 m+1)}{2^{2}(m+1)^{2}}$

$\Rightarrow \frac{(2 m+2) !}{2^{2 m+2}((m+1) !)^{2}} \leq \frac{2 m+1}{2(m+1) \sqrt{3 m+1}}$

$\Rightarrow \frac{(2 m+2) !}{2^{2 m+2}((m+1) !)^{2}} \leq \sqrt{\frac{(2 m+1)^{2}}{4(m+1)^{2}(3 m+1)}}$

$\Rightarrow \frac{(2 m+2) !}{2^{2 m+2}((m+1) !)^{2}} \leq \sqrt{\frac{\left(4 m^{2}+4 m+1\right) \times(3 m+4)}{4\left(3 m^{3}+7 m^{2}+5 m+1\right)(3 m+4)}}$

$\Rightarrow \frac{(2 m+2) !}{2^{2 m+2}((m+1) !)^{2}} \leq \sqrt{\frac{12 m^{3}+28 m^{2}+19 m+4}{\left(12 m^{3}+28 m^{2}+20 m+4\right)(3 m+4)}}$

$\because \frac{12 m^{3}+28 m^{2}+19 m+4}{\left(12 m^{3}+28 m^{2}+20 m+4\right)}<1$

$\therefore \frac{(2 m+2) !}{2^{2 m+2}((m+1) !)^{2}}<\frac{1}{\sqrt{3 m+4}}$

Thus, $P(m+1)$ is true.

Hence, by mathematical induction $\frac{(2 n) !}{2^{2 n}(n !)^{2}} \leq \frac{1}{\sqrt{3 n+1}}$ is true for all $n \in N$