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Math Help - Recurrence relation proof

  1. #1
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    Recurrence relation proof

    n = 1,2,3...

    I_{n} = \int \frac {x^{n-1}}{2-x} dx between limits 0 and 1.

    Writing x^{n} = x^{n-1}(2-(2-x)), show this sequence satisfies I_{n+1} = 2I_{n} - \frac {1}{n}.

    Now, I can't quite get a foothold into this puzzle. I'm thinking there's a simple logical step but I don't see it. Can someone help please? I wanted to try induction but I don't see how I'm supposed to use the idea they've told me to do so. It is required of the problem to use this hint.
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  2. #2
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    Re: Recurrence relation proof

    Use the hint

    I_{n+1} = \int_0^1 \frac{x^n}{2-x}\,dx = \int_0^1 \frac{x^{n-1}\left(2 - (2-x)\right)}{2-x}\,dx = \int_0^1 \frac{2x^{n-1}}{2-x}\,dx - \int_0^1 x^{n-1} \frac{2-x}{2-x}\,dx

    so

    I_{n+1} = 2 I_{n} - \int_0^1 x^{n-1}\,dx.

    Now integrate the last term.
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