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Question: 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 +...

If I(m, n) = ∫01tm(1+t)ndt\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

A

2nm+1−nm+1I(m+1,n−1)\frac{2^{n}}{m + 1} - \frac{n}{m + 1}I(m + 1,n - 1)

B

nm+1I(m+1,n−1)\frac{n}{m + 1}I(m + 1,n - 1)

C

2nm+1+nm+1I(m+1,n−1)\frac{2^{n}}{m + 1} + \frac{n}{m + 1}I(m + 1,n - 1)

D

mn+1I(m+1,n−1)\frac{m}{n + 1}I(m + 1,n - 1)

Answer

2nm+1−nm+1I(m+1,n−1)\frac{2^{n}}{m + 1} - \frac{n}{m + 1}I(m + 1,n - 1)

Explanation

Solution

I(m , n) = ∫01tm(1+t)ndt\int_{0}^{1}{t^{m}(1 + t)^{n}dt} = (1 + t)n

 tm+1m+1∣01−∫01n(1+t)n−1tm+1m+1dt\left. \ \frac{t^{m + 1}}{m + 1} \right|_{0}^{1} - \int_{0}^{1}{n(1 + t)^{n - 1}\frac{t^{m + 1}}{m + 1}dt}

= 2nm+1−nm+1I(m+1,n−1)\frac{2^{n}}{m + 1} - \frac{n}{m + 1}I(m + 1,n - 1).