Question

Coefficient of $x^3$ in the expansion of ${\left(\frac{1+x}{1-x}\right)}^2$, |x| < 1 is

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Answer 1

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

We will expand ${(1-x)}^{-2}$ and find the coefficient of $x^3$

${(1+x)}^2$${(1-x)}^{-2}$ = (1 + 2x + $x^2$)(1 + 2x + 3$x^2$ + 4$x^3$...........)

[${(1+x)}^n$ = 1 + nx + $\frac{n(n-1)}{2!}$$x^2$ + $\frac{n(n-1)(n-3)}{3!}$$x^3$..........]

= 1(1 + 2x + 3$x^2$ + 4$x^3$..........) + 2x(1 + 2x + 3$x^2$ + 4$x^3$..........) + $x^2$(1 + 2x + 3$x^2$ + 4$x^3$..........)

Coefficient of $x^3$ = 4 + 6 + 2 = 12