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7.Binomial Theorem
normal
Coefficient of $x^6$ in the binomial expansion ${\left( {\frac{{4{x^2}}}{3}\; - \;\frac{3}{{2x}}} \right)^9}$ is
A
$2438$
B
$2688$
C
$2868$
D
none
Solution
$\left(\frac{4 x^{2}}{3}-\frac{3}{2 x}\right)^{9}$
$T_{r+1}={ }^{9} C_{r}\left(\frac{4 x^{2}}{3}\right)^{9-f}\left(-\frac{3}{2 x}\right)^{r}$
Power of $x=2(9-r)+(-1) r$
$\quad=18-3 r=6$
$\Rightarrow 3 r=18-6=12 \Rightarrow r=4$
Coefficient ${ }^{9} C _{4}\left(\frac{4}{3}\right)^{9-4}\left(\frac{-3}{2}\right)^{4}$
$=\frac{9 \times 8 \times 7 \times 6}{1.2 .3 .4}\left(\frac{4}{3}\right)^{5}\left(\frac{3}{2}\right)^{4}$
$=9 \times 2 \times 7 \times \frac{2^{10} \times 3^{4}}{3^{5} \times 2^{4}}$
$=21 \times 2^{7}=2688$
Standard 11
Mathematics
