A) 2002
B) \[\frac{{{(2002)}^{2001}}}{(2001)!}\]
C) \[\frac{{{(2001)}^{2002}}}{(2002)!}\]
D) 2001
Correct Answer: B
Solution :
| [b] \[\frac{\pi (n)}{\pi (n+1)}=\frac{^{n}{{C}_{0}}{{.}^{n}}{{C}_{1}}{{.}^{n}}{{C}_{2}}{{...}^{n}}{{C}_{n}}}{^{n+1}{{C}_{0}}{{.}^{n+1}}{{C}_{1}}{{.}^{n+1}}{{C}_{2}}{{...}^{n+1}}{{C}_{n+1}}}\] |
| \[=\frac{1}{^{n+1}{{C}_{0}}}\left( \frac{^{n}{{C}_{0}}}{^{n+1}{{C}_{1}}} \right)\left( \frac{^{n}{{C}_{1}}}{^{n+1}{{C}_{2}}} \right).....\left( \frac{^{n}{{C}_{n}}}{^{n+1}{{C}_{n+1}}} \right)\] |
| \[=\frac{1}{1}\left( \frac{1}{n+1} \right)\left( \frac{2}{n+1} \right)......\left( \frac{n+1}{n+1} \right)\] |
| \[\left[ \because \,\frac{^{n}{{C}_{r}}}{^{n+1}{{C}_{r+1}}}=\frac{r+1}{n+1} \right]\] |
| \[=\frac{(n+1)!}{{{(n+1)}^{n+1}}}=\frac{n!}{{{(n+1)}^{n}}}\] |
| \[\therefore \,\,\,\,\,\frac{\pi (2002)}{\pi (2001)}=\frac{{{(2002)}^{2001}}}{(2001)!}\] |
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