If $I(m, n) = \int\limits_0^1t^m(1+t)^ndt$, then the expression for I(m, n) in terms of I(m + 1, n – 1) is
Answer & explanation
Correct answer: option 1
$I(m, n) = \int\limits_0^1t^m(1+t)^n\,dt$
$=(1+t)^n\frac{t^{m+1}}{m+1}|_0^1-\int\limits_0^1n(1+t)^{n-1}\frac{t^{m+1}}{m+1}$
$=\frac{2^n}{m+1}-\frac{n}{m+1}\,I(m+1,1-1)$
Hence (A) is the correct answer.