A) \[\frac{{{(-1)}^{n}}n!}{{{a}^{n+1}}}\]
B) \[\frac{{{(-1)}^{n}}(n-1)!}{{{a}^{n}}}\]
C) \[\frac{n!}{{{a}^{n+1}}}\]
D) None of these
Correct Answer: C
Solution :
[c] Let \[{{I}_{n}}=\int\limits_{0}^{\infty }{{{x}^{n}}{{e}^{-ax}}=\left[ {{x}^{n}}.\frac{{{e}^{-ax}}}{-a} \right]_{0}^{\infty }-\int\limits_{0}^{\infty }{n{{x}^{n-1}}.\frac{{{e}^{-ax}}}{-a}dx}}\] \[=-\frac{1}{a}\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{n}}}{{{e}^{ax}}}+\frac{n}{a}{{I}_{n-1}}\therefore {{I}_{n}}=\frac{n}{a}{{I}_{n-1}}\] \[=\frac{n}{a}\cdot \frac{n-1}{a}{{I}_{n-2}}=\frac{n(n-1)(n-2)}{{{a}^{3}}}{{I}_{n-3}}\] \[=\frac{n!}{{{a}^{n}}}\int\limits_{0}^{\infty }{{{e}^{-ax}}dx=\frac{n!}{{{a}^{n+1}}}}\]
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