A) \[\frac{n+1}{n!}\]
B) \[\frac{{{n}^{n}}}{n!}\]
C) \[\frac{{{(n+1)}^{n}}}{(n+1)!}\]
D) \[\frac{{{(n+1)}^{n+1}}}{(n+1)!}\]
Correct Answer: D
Solution :
[d] Here, \[{{P}_{n}}={{\,}^{n}}{{C}_{0}}.{{\,}^{n}}{{C}_{1}}.{{\,}^{n}}{{C}_{2}}....{{\,}^{n}}{{C}_{n}}\] and \[{{P}_{n+1}}={{\,}^{n+1}}{{C}_{0}}.{{\,}^{n+1}}{{C}_{1}}.{{\,}^{n+1}}{{C}_{2}}....{{\,}^{n+1}}{{C}_{n+1}}\] \[\therefore \,\frac{{{P}_{n+1}}}{{{P}_{n}}}=\frac{^{n+1}{{C}_{0}}.{{\,}^{n+1}}{{C}_{1}}.{{\,}^{n+2}}{{C}_{2}}....{{\,}^{n+1}}{{C}_{n+1}}}{^{n}{{C}_{0}}.{{\,}^{n}}{{C}_{1}}.{{\,}^{n}}{{C}_{2}}....{{\,}^{n}}{{C}_{n}}}\] \[=\left( \frac{^{n+1}{{C}_{1}}}{^{n}{{C}_{0}}} \right)\left( \frac{^{n+1}{{C}_{3}}}{^{n}{{C}_{1}}} \right)\left( \frac{^{n+1}{{C}_{3}}}{^{n}{{C}_{0}}} \right)....\left( \frac{^{n+1}{{C}_{n+1}}}{^{n}{{C}_{n}}} \right)\] \[=\left( \frac{n+1}{1} \right)\left( \frac{n+1}{2} \right)\left( \frac{n+1}{3} \right)....\left( \frac{n+1}{n+1} \right)\] \[=\frac{{{(n+1)}^{n+1}}}{(n+1)!}\]You need to login to perform this action.
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