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7.Binomial Theorem
easy
The coefficient of ${x^{ - 7}}$ in the expansion of ${\left( {ax - \frac{1}{{b{x^2}}}} \right)^{11}}$ will be
A
$\frac{{462{a^6}}}{{{b^5}}}$
B
$\frac{{462{a^5}}}{{{b^6}}}$
C
$\frac{{ - 462{a^5}}}{{{b^6}}}$
D
$\frac{{ - 462{a^6}}}{{{b^5}}}$
(IIT-1967)
Solution
(b) For number of term, $(11 – r)(1) + r( – 2) = – 7$
$\Rightarrow 11 – r – 2r = – 7$
$ \Rightarrow r = 6$
Thus coefficient of $x^{-7}$ is $^{11}{C_6}{(a)^5}{\left( { – \frac{1}{b}} \right)^6} $
$= \frac{{462}}{{{b^6}}}{a^5}$
Standard 11
Mathematics
