If `I(m,n)=int_0^1 t^m(1+t)^n.dt`, then the expression for `I(m,n)` in terms of `I(m+1,n-1)` is:

Options

A. `(2^(n))/(m+1)-(n)/(m+1)I(m+1,n-1)`
B. `(n)/(m+1)I(m+1,n-1)`
C. `(2^(n))/(m+1)-(n)/(m+1)I(m+1,n-1)`
D. `(m)/(n+1)I(m+1,n-1)`

Answer

B

Explanation

We have,
`I(m,n)-overset(1)underset(0)intt^(m)(1-t)^(n)dt`
`rArr 1(m+1,n-1)=overset(1)underset(0)int t^(m+1)(1-t)^(n-1)dt`
`rArr I(m+1,n-1)=[-(t^(m+1)(1-t)^(n))/(n)]_(0)^(1)+(m+1)/(n)underset(0)overset(1)int t^(m)(1-t)^(n)dt`
`rArr 1(m+1,n-1)=(m+1)/(n)I(m,n)`
`rArr (m,n)=(n)/(m+1)I(m+1,n-1)`