In the expansion of , x 0, if the term independent of x is (221)k, then k is equal to: [2026]

Question

In the expansion of ${(9 x - \frac{1}{3 \sqrt{x}})}^{18}$ , x > 0, if the term independent of x is (221)k, then k is equal to: [2026]

Smart Achievers · Accepted Answer

(1)

We have, ${(9 x - \frac{1}{3 \sqrt{x}})}^{18}$

General Term, $T_{r + 1} = {}^{18}C_{r} {(9 x)}^{18 - r} {(\frac{- 1}{3 \sqrt{x}})}^{r}$

For term independent of x, $1 - r - \frac{r}{2} = 0$

$\Rightarrow T_{13} = 18 \Rightarrow r = 12$

$∴ T_{13} = {}^{18}C_{12} {(9 x)}^{6} {(- \frac{1}{3 \sqrt{x}})}^{12}$

$= {}^{18}C_{6} {\times 9}^{6} \times {(\frac{1}{3})}^{12}$

$= \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{720} \times 9^{6} \times \frac{1}{9^{6}}$

= 18564 = 221k (Given)

$∴$ k = 84

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