Question

Find the coefficient of
$x^5$ in the expansion of $(1-3x+3x^2{)}^{-\frac{1}{2}}$.

Answer 1

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

$=1-\frac{1}{2}\left[3x\right(x-1\left)\right]+\frac{3}{8}\left[3x\right(x-1){]}^2-\frac{5}{16}[3x(x-1){]}^3+\frac{35}{128}\left[3x\right(x-1){]}^4-\frac{63}{256}[3x(x-1){]}^5......$

$=1-\frac{1}{2}3x(x-1)+\frac{3}{8}9x^2(x-1{)}^2-\frac{5}{16}27x^3(x-1{)}^3+\frac{35}{128}81x^4(x-1{)}^4-\frac{63}{256}243x^5(x-1{)}^5......$

Now we have to find $x^5$ coefficient in the above expression

To do so we have to check by expanding $(x-1{)}^n$ untill first 6 terms ,since after that the power of x is always greater than 5.

Thus coefficient of $x^5=\frac{-5}{16}.27.(-3)+\frac{35}{128}.81.(-4)+\frac{-63}{256}.\left(243\right).(-1)$

$=\frac{-891}{256}$