First slide

Binomial theorem for positive integral Index

Question

$The 9th term of the expansion of {(3 x - \frac{1}{2 x})}^{8} is$

Easy

A

$\frac{1}{512 x^{9}}$
B

$\frac{- 1}{512 x^{9}}$
C

$\frac{- 1}{256 x^{8}}$
D

$\frac{1}{256 x^{8}}$

Solution

${(3 x - \frac{1}{2 x})}^{8}$
$When we expand the given binomial, we get 9 terms, 9 th term is the last term. T_{r + 1} = {}^{n}C_{r} \cdot x^{n - r} a^{r} . for (x + a)^{r} T_{9} = {}^{8}C_{8} (3 x)^{0} {(\frac{- 1}{2 x})}^{8} = \frac{+ 1}{256 x^{8}}$

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