Prove by induction that $I_n = \frac{4^{n+1}n!(n+1)!}{(2n+3)!}$

Question

$I_n$ is defined as:

$$I_n = \int_{0}^{1} \big[x^n \sqrt{1-x}\big] dx$$

Let $p(n)$ be the statement:

$$I_n = \frac{4^{n+1}n!(n+1)!}{(2n+3)!}$$

Prove by mathematical induction $p(n)$ is true for n = 0,1,2,...

It is straightforward to show that the statement $p(0)$ is true and $I_0 = \tfrac{2}{3}$

However I do not no where to begin to prove that $p(k) \Rightarrow p(k+1)$ and trying to integrate $I_n$ is getting me nowhere.

I am not necessarily asking for a general proof, rather some hints about how to tackle it, and whether I need to solve the integration statement.

Answer 1

Hint: Write $x^n\sqrt{1-x}$ as $x^{n-1}(1-(1-x))\sqrt{1-x}$. That gives: $$ I(n) = I(n-1) - \int_{0}^{1} x^{n-1}(1-x)^{3/2}\,dx. $$ Use integration by parts on the last integral, and it will reveal itself as a peculiar multiple of $I(n)$.

The previous relation can so be rearranged in order to give a closed formula for $\frac{I(n)}{I(n-1)}$ in terms of $n$ only. Compute $I(0)$ and profit by induction.

Answer 2

Let $I_n$ be the integral defined by

$$\begin{align} I_n&=\int_0^1 x^n\sqrt{1-x}\,dx\\\\ &=\int_0^1 (1-x)^n \sqrt{x}\,dx \end{align}$$

Integrating by parts with $u=(1-x)^n$ and $v=\frac23 x^{3/2}$ reveals

$$\begin{align} I_n&=\frac23 n\int_0^1 (1-x)^{n-1}\,x^{3/2}\,dx\\\\ &=\frac23 n I_{n-1}-\frac23 nI_n\\\\ &=\frac{\frac23 n}{1+\frac23 n}I_{n-1}\\\\ &=\frac{2n}{2n+3}I_{n-1} \tag 1 \end{align}$$

Using $I_0=\frac23$ along with $(1)$ and proceeding recursively, we obtain

$$\begin{align} I_n&=\frac{2^{n+1}\,n!}{(2n+3)!!}\\\\ &=\frac{2^{n+1}\,n!}{\frac{(2n+3)!}{2^{n+1}\,(n+1)!}}\\\\ &=\frac{4^{n+1}\,n!\,(n+1)!}{(2n+3)!} \end{align}$$

as was to be shown!