Binomial Theorem And Its Simple Applications Question 41
Question: The coefficient of $ {x^{-7}} $ in the expansion of $ {{[ ax-\frac{1}{bx^{2}} ]}^{11}} $ will be:
Options:
A) $ \frac{462}{b^{5}}a^{6} $
B) $ \frac{462a^{5}}{b^{6}} $
C) $ \frac{-462a^{5}}{b^{6}} $
D) $ \frac{-462a^{6}}{b^{5}} $
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Answer:
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}{bx^{2}} $
$ \therefore ,{T_{r+1}}={{,}^{11}}C_{r}{{(ax)}^{11-r}}{{( \frac{-1}{bx^{2}} )}^{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}}{{( \frac{-1}{b} )}^{6}}=\frac{462}{b^{2}}a^{5} $
BETA
