A) \[\frac{(2n)!}{{{(n!)}^{2}}}\]
B) \[\frac{2n}{{{(n!)}^{2}}}\]
C) \[\frac{(2n)!}{{{n}^{2}}}\]
D) None of these
Correct Answer: A
Solution :
General term in the expansion of\[{{(1-4x)}^{1/2}}\]is \[=\frac{\left( -\frac{1}{2} \right)\left( -\frac{1}{2}-1 \right)\left( -\frac{1}{2}-2 \right)....\left\{ -\frac{1}{2}-(r-1) \right\}}{r!}\] \[{{(-4x)}^{r}}\] \[\Rightarrow \]\[{{T}_{r+1}}=\frac{{{(-1)}^{r}}\left( \frac{1}{2} \right)\left( \frac{3}{2} \right)\left( \frac{5}{2} \right)....\left( \frac{2r-1}{2} \right)}{r!}\] \[{{(-1)}^{r}}{{4}^{r}}{{x}^{r}}\] \[\Rightarrow \]\[{{T}_{r+1}}={{(-1)}^{2r}}\frac{1.3.5.7....(2r-1)}{{{2}^{r}}r!}{{2}^{2r}}{{x}^{r}}\] \[\Rightarrow \]\[{{T}_{r+1}}=\frac{1.3.5.7....(2r-1)}{r!}{{2}^{r}}.{{x}^{r}}\] ?.(i) For the coefficient of\[{{x}^{n}},\]put\[r=n\] in Eq. (i), we get \[{{T}_{n+1}}=\frac{1.3.5.7....(2n-1)}{n!}{{.2}^{n}}.{{x}^{n}}\] Coefficient of\[{{x}^{n}}\] \[=\frac{1.3.5.7...(2n-1)}{n!}{{.2}^{n}}\] \[=\frac{1.2.3.4.5.6....(2n-2)(2n-1)(2n)}{2.4.6....(2n-2)(2n).n!}{{.2}^{n}}\] \[=\frac{(2n)!{{.2}^{n}}}{{{2}^{n}}[1.2.3....(n-1)n]n!}\] \[=\frac{(2n)!}{n!n!}=\frac{(2n)!}{{{(n!)}^{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
