Question

If $l(m,n)={\int }_0^1{t}^m(1+t{)}^{n}dt$, then the expression for $l(m,n)$ in terms of $l(m+1,n-1)$ is

• A. $\frac{2^n}{m+1}-\frac{n}{m+1}.l(m+1,n-1)$
• B. $\frac{n}{m+1}.l(m+1,n-1)$
• C. $\frac{2n}{m+1}+\frac{n}{m+1}l.(m+1,n-1)$
• D. $\frac{m}{n+1}.l(m+1,n-1)$

Answer 1

Answer: (A)

$\frac{2^n}{m+1}-\frac{n}{m+1}.l(m+1,n-1)$

Here, $I(m,n)={\int }_0^1{t}^m{(1+t)}^{n}dt$ reduce into $I(m+1,n-1)$

We have, $I(m+1)(n-1)={\int }_0^1{t}^{m+1}(1+t{)}^{n-1}$

(We apply integration by parts taking ${(1+t)}^n$ as first and $t^m$ as second function)

$\therefore I(m,n)={[{(1+t)}^n.\frac{t^{m+1}}{m+1}]}_0^1-{\int }_0^{1}n{(1+t)}^{(n-1)}.\frac{t^{m+1}}{m+1}dt$

$\therefore I(m,n)=[{\left(2\right)}^n.\frac{1^{(m+1)}}{m+1}-0]-{\int }_0^{1}n{(1+t)}^{(n-1)}.\frac{t^{m+1}}{m+1}dt$

$\therefore I(m,n)=\frac{2^n}{m+1}-\frac{n}{m+1}{\int }_0^1{(1+t)}^{(n-1)}.t^{m+1}dt$

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