Binomial TheoremMedium
Question
In the expansion of $\left( 9x - \frac{1}{3\sqrt{x}} \right)^{18},x > 0$, if the term independent of x is $(221)k$. then k is equal to:
Options
A.84
B.78
C.168
D.198
Solution
Sol. [Key Concept: In $T_{r + 1}$ set the power of $x$ to $0$ (gives $r = 12$); the resulting coefficient equals $221k$.]
General term in above expression is
$${T_{r + 1} = \ ^{18}C_{r}(9x)^{18 - r}\left( - \frac{1}{3\sqrt{x}} \right)^{r} }{= \left( - \frac{1}{3} \right)^{r}\ ^{18}C_{r}9^{18 - r} \cdot x^{18 - \frac{3r}{2}} }$$For term independent of $x,18 - \frac{3r}{2} = 0 \Rightarrow r = 12$
∴ Coefficient of term independent of x
$${= \left( - \frac{1}{3} \right)^{12}\ ^{18}C_{12}9^{18 - 12} }{= 18564 }$$$= 221k$ (given)
$$\Rightarrow k = 84 $$
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