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Options

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

Answer

A

Explanation

Here , `I(m,n)=int_(0)^(1)t^(m)(1+t)^(n)dt` reduce into ` I (m=1 , n -1)` [ we apply integration by parts taking `(1+t)^(n)` as first and `t^(m)` as second function]
`I(m , n) = [ (1+t)^(n)*(t^(m+1))/(m+1)]_(0)^(1)- int_(0)^(1) n (1+t)^((n-1))*(t^(m+1))/(m+1)dt`
`= (2^(n))/(m+1)-(n)/(m+1)int_(0)^(1)(1+t)^((n-1))*t^(m+1)dt`
`:. I (m , n) = (2^(n))/(m+1)-(n)/(m+1)*I(m+1,n-1)`