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If L(m,n)=int(0)^(1)t^(m)(1+t)^(n),dt, t...

If `L(m,n)=int_(0)^(1)t^(m)(1+t)^(n),dt`, then prove that
`L(m,n)=(2^(n))/(m+1)-n/(m+1)L(m+1,n-1)`

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The correct Answer is:
NA
`L(m,n)=int_(0)^(1)t^(m)(1+t)^(n)dx`
`=[(t^(m+1))/(m+1)(1+t)^(n)]_(0)^(1)-int_(0)^(1)n(1+t)^(n-1)(t^(m+1))/(m+1)dx`
`=[(t^(m+1))/(m+1)(1+t^(n))]_(0)^(1)-n/(m+1)int_(0)^(1)t^(m+1)(1+t)^(n-1)dx`
`(2^(n))/(m+1)-n/(m+1)L(m+1,n-1)`
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