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Expansions of the Form (1+x)^(-n) and (1-x)^(-n)

Coeffiecient ...

Question

Coeffiecient of $x^{4}$ in $\frac{(1 + x)^{2}}{(1 - x)^{3}}$ is
$(| x | < 1)$

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Solution

$\frac{(1 + x)^{2}}{(1 - x)^{3}} = (1 + x)^{2} (1 - x)^{- 3}$
$= (1 + x^{2} + 2 x) (1 - x)^{- 3}$

We know that coefficient of $x^{r}$ in $(1 - x)^{- n} =^{n + r - 1} C_{r}$
$∴ x^{4}$ coefficient
$= 1 \times x^{4}$ coefficient in $(1 - x)^{- 3} + 2 \times x^{3}$ coefficient in $(1 - x)^{- 3} + 1 \times x^{2}$ coefficient in $(1 - x)^{- 3}$
$= 1 \times^{3 + 4 - 1} C_{4} + 2 \times^{3 + 3 - 1} C_{3} + 1 \times^{3 + 2 - 1} C_{2}$
$=^{6} C_{4} + 2 \times^{5} C_{3} +^{4} C_{2}$
$= 15 + 20 + 6$
$= 41$

Alternate Solution:
$\frac{(1 + x)^{2}}{(1 - x)^{3}} = (1 + x)^{2} (1 - x)^{- 3} = (1 + 2 x + x^{2}) (1 + 3 x + 6 x^{2} + 10 x^{3} + 15 x^{4} + \dots)$
Hence $x^{4}$ coefficient will be
$= 1 \cdot 15 + 2 \cdot 10 + 1 \cdot 6 = 41$

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