Problem source

Prove that, for any integers $n$ and $r$, with $1\leqslant r\leqslant n,$
\[
\binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}.
\]
Hence or otherwise, prove that 
\[
(uv)^{(n)}=u^{(n)}v+\binom{n}{1}u^{(n-1)}v^{(1)}+\binom{n}{2}u^{(n-2)}v^{(2)}+\cdots+uv^{(n)},
\]
where $u$ and $v$ are functions of $x$ and $z^{(r)}$ means $\dfrac{\mathrm{d}^{r}z}{\mathrm{d}x^{r}}$. 
Prove that, if $y=\sin^{-1}x,$ then $(1-x^{2})y^{(n+2)}-(2n+1)xy^{(n+1)}-n^{2}y^{(n)}=0.$

Solution source

\begin{align*}
\binom{n}{r} + \binom{n}{r-1} &= \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} \\
&= \frac{n!}{(r-1)!(n-r)!} \left ( \frac{1}{r} + \frac{1}{n-r+1} \right) \\
&= \frac{n!}{(r-1)!(n-r)!} \frac{(n-r+1)+r}{r(n-r+1)} \\
&= \frac{n! (n+1)}{r! (n-r+1)!} \\
&= \frac{(n+1)!}{r!(n+1-r)!} \\
&= \binom{n+1}{r}
\end{align*}

Claim: $\displaystyle (uv)^{(n)} = \sum_{r=0}^n \binom{n}{r} u^{(n-r)} v^{(r)}$

Proof: (By induction on $n$).

Base case: $n = 0$ is clear.

Inductive step: Suppose it is true for $n = k$, then consider

\begin{align*}
(uv)^{(k+1)} &= \left ( (uv)^{(k)} \right)' \\
&= \left (  \sum_{r=0}^k \binom{k}{r} u^{(k-r)} v^{(r)} \right)' \tag{by assumption} \\
&=\sum_{r=0}^k \binom{k}{r}  \left (  u^{(k-r)} v^{(r)}\right)'  \tag{linearity} \\
&=\sum_{r=0}^k \binom{k}{r}  \left (  u^{(k-r+1)} v^{(r)} + u^{(k-r)}v^{(r+1)}\right) \\
&= \sum_{r=0}^{k} \binom{k}{r} u^{(k-r+1)} v^{(r)}  + \sum_{r=0}^{k} \binom{k}{r}  u^{(k-r)}v^{(r+1)} \\
&= \sum_{r=0}^{k} \binom{k}{r} u^{(k-r+1)} v^{(r)}  + \sum_{r=1}^{k+1} \binom{k}{r-1}  u^{(k-r+1)}v^{(r)} \\
&= u^{(k+1)}v + \sum_{r=1}^k \left (\binom{k}{r} + \binom{k}{r-1} \right)u^{(k-r+1)}v^{(r)} + u v^{(k+1)}\\
 &= u^{(k+1)}v + \sum_{r=1}^k \binom{k+1}{r} u^{(k-r+1)}v^{(r)} + u v^{(k+1)}\\
 &= \sum_{r=0}^{k+1} \binom{k+1}{r} u^{(k-r+1)}v^{(r)}\\
\end{align*}

Therefore if our statement is true for $n = k$ it is true for $n = k+1$. Since it is true for $n = 0$ by the principle of mathematical induction it is true for all integer $n \geq 0$

Suppose $ y = \sin^{-1} x$, then $y' = \frac{1}{\sqrt{1-x^2}}$, $y'' = \frac{x}{(1-x^2)^{3/2}}$.

Not that this means that $(1-x^2)y'' - xy' = 0$ (which is our formula when $n = 0$).

Now apply Leibniz's formula to this.

\begin{align*}
0 &= \left ( (1-x^2)y'' - xy'  \right)^{(n)} \\
&= \left ( (1-x^2)y'' \right)^{(n)} -\left ( xy'  \right)^{(n)} \\
&= \left ( (1-x^2)y^{(n+2)} - n\cdot 2x \cdot y^{(n+1)}-\binom{n}{2} \cdot 2 \cdot y^{(n)} \right )- \left (xy^{(n+1)}+ny^{(n)} \right) \\
&= (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} - \left ( n(n-1)+n \right)y^{(n)} \\
&= (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} - n^2y^{(n)} \\
\end{align*}

as required