Binomial Theorem And Its Simple Applications Question 191
Question: If $ S_{n}=\sum\limits_{r=0}^{n}{\frac{1}{^{n}C_{r}}} $ and $ t_{n}=\sum\limits_{r=0}^{n}{\frac{r}{^{n}C_{r}}} $ , then $ \frac{t_{n}}{S_{n}} $ is equal to
[AIEEE 2004]
Options:
A) $ \frac{2n-1}{2} $
B) $ \frac{1}{2}n-1 $
C) $ n-1 $
D) $ \frac{1}{2}n $
Show Answer
Answer:
Correct Answer: D
Solution:
- We have, $ S_{n}=\sum\limits_{r=0}^{n}{\frac{1}{^{n}C_{r}}} $ and $ t_{n}=\sum\limits_{r=0}^{n}{\frac{r}{^{n}C_{r}}} $
$ t_{n}=\sum\limits_{r=0}^{n}{\frac{n-(n-r)}{^{n}{C_{n-r}}}} $ , $ [\because ,{{,}^{n}}C_{r}={{,}^{n}}{C_{n-r}}] $ = $ n\sum\limits_{r=0}^{n}{\frac{1}{^{n}C_{r}}}-\sum\limits_{r=0}^{n}{\frac{n-r}{^{n}{C_{n-r}}}} $
$ t_{n}=n,.,S_{n}-[ \frac{n}{^{n}C_{n}}+\frac{n-1}{^{n}{C_{n-1}}}+…..+\frac{1}{^{n}C_1}+0 ] $
$ t_{n}=n,.,S_{n}-\sum\limits_{r=0}^{n}{\frac{r}{^{n}C_{r}}} $
$ \Rightarrow $ $ t_{n}=n,.,S_{n}-t_{n} $
$ \Rightarrow $ $ 2t_{n}={{,}^{n}}S_{n}\Rightarrow \frac{t_{n}}{S_{n}}=\frac{n}{2} $ .
BETA
