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 $

Question

I would appreciate if somebody could help me with the following problem:

Please explain how to do this proof ? $$$$ If $$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 .$$

score 1 · Answer 1 · answered Feb 22 '13 at 14:48

1

Hint: Try integration by parts twice.

answered Feb 22 '13 at 14:48

EuYu

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Can the OP use the Beta unction here? Is it applicable here. – Mikasa Feb 22 '13 at 14:52
@BabakS. I might be mistaken, but I don't think you need the Beta function here. – EuYu Feb 22 '13 at 14:57

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 $

1 Answers1