A) \[\frac{{{2}^{n}}}{m+1}-\frac{n}{m+1}I(m+1,\,\,n-1)\]
B) \[\frac{n}{m+1}I(m+1,\,\,n-1)\]
C) \[\frac{{{2}^{n}}}{m+1}+\frac{n}{m+1}I(m+1,\,\,n-1)\]
D) \[\frac{m}{m+1}I(m+1,\,\,n-1)\]
Correct Answer: B
Solution :
Here,\[I(m,\,\,n)=\int_{0}^{1}{{{t}^{m}}}{{(1+t)}^{n}}dt\] We apply integration by parts, taking\[{{(1+t)}^{\pi }}\]as first and\[{{t}^{m}}\]as second function] \[I(m,\,\,n)=\left[ {{(1+t)}^{n}}\cdot \frac{{{t}^{m+1}}}{m+1} \right]_{0}^{1}\] \[-\int_{0}^{1}{n{{(1+t)}^{n-1}}\cdot \frac{{{t}^{m+1}}}{m+1}dt}\] \[=\frac{{{2}^{n}}}{m+1}-\frac{n}{m+1}\int_{0}^{1}{{{(1+t)}^{n-1}}\cdot {{t}^{m+1}}dt}\] \[\therefore \] \[I(m,\,\,n)=\frac{{{2}^{n}}}{m+1}-\frac{n}{m+1}\cdot I(m+1,\,\,n-1)\]
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