Q78 / 120 · VITEEE 2014 · Mathematics
If $l(m, n)=\int_{0}^{1} t^{m}(1+t)^{n} d t$, then the expression for $l(m, n)$ in terms of $l(m+1, n+1)$ is
- Correct answerA$\frac{2^{n}}{m+1}-\frac{n}{m+1} \cdot l(m+1, n-1)$
- B$\frac{n}{m+1} \cdot l(m+1, n-1)$
- C$\frac{2 n}{m+1}+\frac{n}{m+1} l .(m+1, n-1)$
- D$\frac{m}{n+1} \cdot l(m+1, n-1)$
Solution
$\begin{aligned} & l(\mathrm{~m}, \mathrm{n})=l=\int_{0}^{1} \mathrm{t}^{\mathrm{m}}(1+\mathrm{t})^{\mathrm{n}} \mathrm{dt} \\ \Rightarrow & l(\mathrm{~m}, \mathrm{n})=\left[(1+\mathrm{t})^{\mathrm{n}} \cdot \frac{\mathrm{t}^{\mathrm{m}+1}}{\mathrm{~m}+1}\right]_{0}^{1} \\ &-\frac{\mathrm{n}}{\mathrm{m}+1} \int_{0}^{1}(1+\mathrm{t})^{\mathrm{n}-1} \cdot \mathrm{t}^{\mathrm{m}+1} \cdot \mathrm{dt} \\=& \frac{2^{\mathrm{n}}}{\mathrm{m}+1}-\frac{\mathrm{n}}{\mathrm{m}+1} \cdot l(\mathrm{~m}+1, \mathrm{n}-1) \end{aligned}$
Hence, the correct answer is option: 'A'.