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Q. If the binomial expansion of $(a + bx)^{-2}$ is $\frac{1}{4} -3x + ...$, then the value of $a$ and $b$ are

Binomial Theorem

Solution:

We have, $\left(a+bx\right)^{-2} = \frac{1}{4}-3x+\ldots$ (Given)
Also, $\left(a+bx\right)^{-2} = a^{-2}\left(1+\frac{b}{a}x\right)^{-2}$
$= a^{-2}\left(1-\frac{2b}{a}x\right)$
[Neglecting $x^{2}$ and higher powers of $x$]
$= \frac{1}{a^{2}}-\frac{2bx}{a^{3}}$
$\therefore \frac{1}{a^{2}}-\frac{2bx}{a^{3}} = \frac{1}{4}-3x$
$\Rightarrow \frac{1}{a^{2}} = \frac{1}{4}$ and $\frac{2b}{a^{3}} = 3$
$\Rightarrow a = 2$ and $b= \frac{3}{2}a^{3} $
$= \frac{3}{2}\times\left(2\right)^{3} = \frac{3}{2} \times8 = 12$