Binomial Theorem And Its Simple Applications Question 13
Question: The value of $ \sum\limits_{r=0}^{50}{{{(-1)}^{r}}} $ $ \frac{^{50}C_{r}}{r+2} $ is equal to
Options:
A) $ \frac{1}{50\times 51} $
B) $ \frac{1}{52\times 50} $
C) $ \frac{1}{52\times 51} $
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Here, $ T_{r}={{(-1)}^{r}}\frac{^{50}C_{r}}{r+2} $
$ ={{(-1)}^{r}}(r+1)\frac{^{50}C_{r}}{(r+1)(r+2)} $
$ ={{(-1)}^{r}}(r+1)\frac{^{52}{C_{r+2}}}{51\times 52} $
$ ={{(-1)}^{r}}\frac{{{[(r+2)-1]}^{52}}{C_{r+2}}}{51\times 52} $
$ ={{(-1)}^{r}}\frac{[52^{51}{C_{r+1}}{{-}^{52}}{C_{r+2}}]}{51\times 52} $
$ =\frac{[-52^{51}{C_{r+1}}{{(-1)}^{r+1}}{{-}^{52}}{C_{r+2}}{{(-1)}^{r+2}}]}{51\times 52} $
$ \sum\limits_{r=0}^{50}{{{(-1)}^{r}}}\frac{^{50}C_{r}}{r+2} $
$ =\sum\limits_{r=0}^{50}{\frac{[-52^{51}{C_{r+1}}{{(-1)}^{r+1}}{{-}^{52}}{C_{r+2}}{{(-1)}^{r+2}}]}{51\times 52}} $
$ =-52\frac{{{(1-1)}^{51}}{{-}^{51}}C_0}{51\times 52}-\frac{{{(1-1)}^{52}}{{-}^{52}}C_0{{+}^{52}}C_1}{51\times 52} $
$ =\frac{1}{51}-\frac{1}{52}=\frac{1}{51\times 52} $ Alternate solution: $ {{(1-x)}^{n}}=\sum\limits_{r=0}^{n}{^{n}C_{r}{{(-1)}^{r}}x^{r}} $ or $ x{{(1-x)}^{n}}=\sum\limits_{r=0}^{n}{{{(-1)}^{r}}{{,}^{n}}}C_{r}{x^{r+1}} $ integrating both sides within the limits 0 to 1, we get $ \int\limits_0^{1}{x{{(1-x)}^{n}}dx}=\sum\limits_{r=0}^{n}{{{(-1)}^{r}}\frac{^{n}C_{r}}{r+2}} $ Or $ \sum\limits_{r=0}^{n}{{{(-1)}^{r}}\frac{^{n}C_{r}}{r+2}=\int\limits_0^{1}{x{{(1-x)}^{n}}dx}} $
$ =\int\limits_0^{1}{(1-x)x^{n}dx} $ (Replace x by 1-x) $ =\frac{{x^{n+1}}}{n+1}-{{. \frac{{x^{n+2}}}{n+2} |}_0}^{1} $
$ =\frac{1}{n+1}-\frac{1}{n+2} $
$ =\frac{1}{(n+1)(n+2)} $ Now put n=50.
$ =\frac{1}{(50)(52)} $
BETA
