The ninth term of the expansion (3x- (1/2x))8 is

Question

The ninth term of the expansion (3x- (1/2x))8 is (A)  (-1/512x9) (B)  (1/512x9) (C)  (1/256.x8) (D)  (-1/512 x8)

Tardigrade Admin · Accepted Answer

The general term of the expansion (x + a)n is Tr+1 =nCr xn-r ar . We have (3x - (1/2x))8 Here, r=8, x =3x, a = (- (1/2x)), n=8 ∴ Nineth term T9 =8C8 (3x)8-8 ((-1/2x))8 = (1/256 .x8)

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

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

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

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

$\frac{- 1}{512 x ^{8}}$

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

Q. The ninth term of the expansion (3x−2x1​)8 is

512x9−1​

512x91​

256.x81​

512x8−1​