First slide

Binomial theorem for positive integral Index

Question

lf the term independent of x in the expansion of ${(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is k, then 18k is equal to

Moderate

A

5
B

7
C

9
D

11

Solution

given binomial is ${(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ So the general term,
$\begin{matrix} T_{r + 1} =^{9} C_{r} {(\frac{3}{2} x^{2})}^{9 - r} {(- \frac{1}{3 x})}^{r} \\ =^{9} C_{r} {(\frac{3}{2})}^{9 - r} {(- \frac{1}{3})}^{r} x^{18 - 3 r} \end{matrix}$
lf the term is independent of x, then $18 - 3 r = 0 \Rightarrow r = 6$
$∴ (r + 1) th$ term =7th term is independent of x
Now, as $T_{6 + 1} = k$
$\begin{array}{l} \Rightarrow^{9} C_{6} {(\frac{3}{2})}^{3} {(- \frac{1}{3})}^{6} = k \\ \Rightarrow \frac{9 \times 8 \times 7}{3 \times 2} \times \frac{1}{27 \times 8} = k \Rightarrow k = \frac{7}{18} \Rightarrow 18 k = 7 \end{array}$
Hence, option (b) is correct.

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