A) \[\frac{462}{{{b}^{5}}}{{a}^{6}}\]
B) \[\frac{462{{a}^{5}}}{{{b}^{6}}}\]
C) \[\frac{-462{{a}^{5}}}{{{b}^{6}}}\]
D) \[\frac{-462{{a}^{6}}}{{{b}^{5}}}\]
Correct Answer: B
Solution :
[b] Suppose \[{{x}^{-7}}\] occurs in \[{{(r+1)}^{th}}\] term. We have \[{{T}_{r+1}}={{\,}^{n}}{{C}_{r}}\,{{x}^{n-r}}\,{{a}^{r}}\] in \[{{(x+a)}^{n}}\]. In the given question, \[n=11,x=ax,\,a=\frac{-1}{b{{x}^{2}}}\] \[\therefore \,{{T}_{r+1}}={{\,}^{11}}{{C}_{r}}{{(ax)}^{11-r}}{{\left( \frac{-1}{b{{x}^{2}}} \right)}^{r}}\] \[={{\,}^{11}}{{C}_{r}}{{a}^{11-r}}\,{{b}^{-r}}{{x}^{11-3r}}{{(-1)}^{r}}\] This term contains \[x{{-}^{7}}\] if \[11-3r=-7\Rightarrow r=6\] Therefore, coefficient of \[{{x}^{-7}}\] is \[^{11}{{C}_{6}}{{(a)}^{5}}{{\left( \frac{-1}{b} \right)}^{6}}=\frac{462}{{{b}^{2}}}{{a}^{5}}\]You need to login to perform this action.
You will be redirected in
3 sec