1) 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: Option AExplanation:$l(m,n)=l=\int_{0}^{1} t^{m}(1+t)^{n} dt$ $\Rightarrow l(m,n)=\left[(1+t)^{n}\frac{t^{m+1}}{m+1}\right]_{0}^{1}$ $-\frac{n}{m+1}\int_{0}^{1} (1+t)^{n-1}.t^{m+1}.dt$ $=\frac{2^{n}}{m+1}-\frac{n}{m+1}l(m+1,n-1)$
