coefficient of $x^9$ in binominal expansion Find the co efficient of $x^9$ in the expansion of $$\left[\frac{(1+x)^3}{(1-x)^3}\right]^\frac{1}{2}$$ I get answer as $\frac{3}{128}$ But in my book the answer is given as $\frac{1}{128}$ I was very much confused .someone help me please.

You could notice that $$\left(\frac{(1+x)^3}{(1-x)^3}\right)^\frac{1}{2}=\left(\frac{1+x}{1-x}\right)^\frac{3}{2}=\left(1+2\sum_{i=1}^\infty x^i\right)^\frac{3}{2}$$ Now, being patient, you could use the generalized binomial theorem.

If you are not, Taylor series (tedious too) would lead to $$\left(\frac{(1+x)^3}{(1-x)^3}\right)^\frac{1}{2}=1+3 x+\frac{9 x^2}{2}+\frac{11 x^3}{2}+\frac{51 x^4}{8}+\frac{57 x^5}{8}+\frac{125 x^6}{16}+\frac{135 x^7}{16}+\frac{1155 x^8}{128}+\frac{1225 x^9}{128}+\frac{2583 x^{10}}{256}+O\left(x^{11}\right)$$ and the result given by Wolfram Alpha, as commented by Arthur.

This coefficient is in fact $\frac {3472875}{9!}=\frac{1225 }{128}$.