Question

If $I(m,n)=\underset{0}{\overset{1}{\int }}{t}^m(1+t{)}^{n}dt$ $(m,n\in N),$ then $I(m,n)$ is:

• A. $\frac{n}{1+m}I\left[\right(m+1),(n-1\left)\right]$
• B. $\frac{m}{m+1}I\left[\right(m+1),(n-1\left)\right]$
• C. $\frac{2^n}{1+m}-\frac{n}{1+m}I\left[\right(m+1),(n-1\left)\right]$
• D. $\frac{2^n}{1+m}-\frac{m}{1+m}I\left[\right(m+1),(n-1\left)\right]$

Answer 1

The correct option is C $\frac{2^n}{1+m}-\frac{n}{1+m}I\left[\right(m+1),(n-1\left)\right]$

$I(m,n)=\underset{0}{\overset{1}{\int }}{\underset{v}{t}}^m\cdot \underset{u}{(1+t{)}^n}dt$
Applying integration by parts
$={\left|\right(1+t{)}^n\frac{t^{m+1}}{m+1}|}_0^1-\underset{0}{\overset{1}{\int }}n(1+t{)}^{n-1}\frac{t^{m+1}}{m+1}dt=\frac{2^n}{m+1}-\frac{n}{m+1}\underset{0}{\overset{1}{\int }}{t}^{m+1}(1+t{)}^{n-1}dt=\frac{2^n}{m+1}-\frac{n}{m+1}I\left[\right(m+1),(n-1\left)\right]$