A) \[\frac{1}{50\times 51}\]
B) \[\frac{1}{52\times 50}\]
C) \[\frac{1}{52\times 51}\]
D) none of these
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) \[=\frac{{{x}^{n+1}}}{n+1}-{{\left. \frac{{{x}^{n+2}}}{n+2} \right|}_{0}}^{1}\] \[=\frac{1}{n+1}-\frac{1}{n+2}\] \[=\frac{1}{(n+1)(n+2)}\] Now put n=50.
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