Mathematics | VITEEE 2014 | Discussion Page

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

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

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

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

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

Option A

$l(m,n)=l=\int_{0}^{1} t^{m}(1+t)^{n} dt$

$\Rightarrow l(m,n)=\left[(1+t)^{n}\frac{t^{m+1}}{m+1}\right]_{0}^{1}$

$-\frac{n}{m+1}\int_{0}^{1} (1+t)^{n-1}.t^{m+1}.dt$

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

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