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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

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

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

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

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

answer is A.

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Detailed Solution

${I (m, n) = \frac{1}{m + 1} t^{m + 1} (1 + t)^{n}]}_{0}^{1}$

$\begin{array}{l} - \frac{n}{m + 1} I (m + 1, n - 1) \\ \Rightarrow I (m, n) = \frac{2^{n}}{m + 1} - \frac{n}{m + 1} I (m + 1, n - 1) \end{array}$

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