Prove If $a_0=2, a_{n}=\frac{\pi^{n+1}}{n!}\int_{0}^{1}t^n(1-t)^n\sin( \pi t)dt(n\geq 1)$ then $a_{n+1}+a_{n-1}=\frac{4n+2}{\pi}a_n $ I would appreciate if somebody could help me with the following problem: 
Please explain how to do this proof ? 
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If
$$a_0=2, a_{n}=\frac{\pi^{n+1}}{n!}\int_{0}^{1}t^n(1-t)^n\sin( \pi t)dt(n\geq 1)~~~$$
then 
$$a_{n+1}+a_{n-1}=\frac{4n+2}{\pi}a_n .$$

Hint: Try integration by parts twice.
