Prove that the coefficient of $x^{3}$ in the expansion of (1 + x + $2 x^{2}$ ) ${(2 x^{2} - \frac{1}{3 x})}^{9} i s - \frac{224}{27} .$

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Solution

$T_{r + 1}$ in ${(2 x^{2} - \frac{1}{3 x})}^{9} =^{9} C_{r} (2 x^{2})^{9 - r} {(- \frac{1}{3 x})}^{r}$
$=^{9} C_{r} 2^{9 - r} (- \frac{1}{3}) x^{18 - 2 r - r}$
This has got to be multiplied by the terms of the first factor i.e., $1 + x + 2 x^{2} = x^{0} + x + 2 x^{2}$
Hence the power of x in (1) after multiplication with the terms of first factor
will be $18 - 3 r, 18 - 3 r + 1, 18 - 3 r + 2$
Each should be equal to 3
$∴ 18 - 3 r = 3 \Rightarrow r = 5$
$∴ 18 - 3 r + 1 = 3$ $∴ r = \frac{16}{3}$ rejected
$∴ 18 - 3 r + 2 = 3$ $∴ r = \frac{17}{3}$ rejected .
Putting r = 5 $T_{r + 1} =^{9} C_{5} 2^{4} {(- \frac{1}{3})}^{5} x^{3}$
= $\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} \cdot (- \frac{16}{243}) x^{3} = - \frac{224}{7} x^{3}$
This when multiplied with (1) of first factor will give the term of $x^{3}$
Hence coefficient of $x^{3}$ in the product is $- \frac{224}{7} .$

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