Question
If $$l\left( {m,\,n} \right) = \int\limits_0^1 {{t^m}{{\left( {1 + t} \right)}^n}} dt,$$ then the expression for $$l\left( {m,\,n} \right)$$ in terms of $$l\left( {m + 1,\,n - 1} \right)$$ is-
A.
$$\frac{{{2^n}}}{{m + 1}} - \frac{n}{{m + 1}}l\left( {m + 1,\,n - 1} \right)$$
B.
$$\frac{n}{{m + 1}}l\left( {m + 1,\,n - 1} \right)$$
C.
$$\frac{{{2^n}}}{{m + 1}} + \frac{n}{{m + 1}}l\left( {m + 1,\,n - 1} \right)$$
D.
$$\frac{m}{{n + 1}}l\left( {m + 1,\,n - 1} \right)$$
Answer :
$$\frac{{{2^n}}}{{m + 1}} - \frac{n}{{m + 1}}l\left( {m + 1,\,n - 1} \right)$$
Solution :
We have $$l\left( {m,\,n} \right) = \int\limits_0^1 {{t^m}{{\left( {1 + t} \right)}^n}} dt$$
Integrating by parts considering $${\left( {1 + t} \right)^n}$$ as first function, we get
$$\eqalign{
& l\left( {m,\,n} \right) = \left[ {\frac{{{t^{m + 1}}}}{{m + 1}}{{\left( {1 + t} \right)}^n}} \right]_0^1 - \frac{n}{{m + 1}}\int\limits_0^1 {{t^{m + 1}}{{\left( {1 + t} \right)}^{n - 1}}dt} \cr
& l\left( {m,\,n} \right) = \frac{{{2^n}}}{{m + 1}} - \frac{n}{{m + 1}}l\left( {m + 1,\,n - 1} \right) \cr} $$