Find the coefficient of $x^{7}$ in the expansion of ${(1 + 3 x - 2 x^{3})}^{10}$ .

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Solution

$[1 + (3 x - 2 x^{3})]^{10}$
General term, $T_{r + 1} =^{10} C_{r} (3 x - 2 x^{3})^{r} =^{10} C_{r} . x^{r} (3 - 2 x^{2})^{r}$
Expansion for $(3 - 2 x^{2})^{r}$ would give eneral term
$T_{y + 1} =^{r} C_{y} (3)^{y} (- 2 x^{2})^{r - y}, 0 \leq y \leq r$
Hence, $T_{r + 1} =^{10} C_{r} . x^{r}^{r} C_{y} {.3}^{y} (- 2)^{r - y} (x^{2})^{r - y}$
$=^{10} C_{r}^{r} C_{y} 3^{y} (- 2 (^{r - y} x^{r + 2 r - 2 y}$
We have, $3 r - 2 y = 7$
As $y \leq r$ , $3 r \geq 7$
$\Rightarrow$ cases: $r = 3, y = 1$ , $r = 4, y = x$
$r = 5, y = 4$ , $r = 6$ ; $r = 7, y = 7$ ; $r = 9, y = 10$
$\Rightarrow$ Coefficient $=^{10} C_{3} \cdot^{3} C_{1} \cdot 3 (- 2)^{2} +^{10} C_{5} \cdot^{5} C_{4} \cdot 3^{4} (- 2) +^{10} C_{7} 3^{7} (- 2)^{o}$ .

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