Question

The ninth term of the expansion $\left(3x-\frac {1}{2x}\right)^8$ is

• A. $\frac {-1}{512x^9}$
• B. $\frac {1}{512x^9}$
• C. $\frac {1}{256.x^8}$
• D. $\frac {-1}{512 x^8}$

Answer 1

Answer: (C)

The general term of the expansion $(x + a)^n$ is
$T_{r+1} =^nC_r x^{n-r} a^r$ .
We have $\left(3x - \frac{1}{2x}\right)^{8} $
Here, $ r=8, x =3x, a = \left(- \frac{1}{2x}\right), n=8 $
$\therefore $ Nineth term $T_{9} =^{8}C_{8} \left(3x\right)^{8-8} \left(\frac{-1}{2x}\right)^{8} $
$= \frac{1}{256 .x^{8}}$