Knowing that:

$\displaystyle (1+x)^k = \sum_{n=0}^{\infty}\binom{k}{n}x^n = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + ... $

Give the Maclaurin series for the function:

$\displaystyle f(x) = \frac{x^2}{\sqrt{2+x}}$

I figure I can rewrite the equation as $\displaystyle f(x) = x^2(1+(1+x))^{-1/2}$ so that my series becomes:

$\displaystyle x^2(1+(x+1))^{-\frac1{2}} = x^2\sum_{n=0}^{\infty}\binom{(-\frac1{2})}{n}(x+1)^n =$$\displaystyle x^2 [1 + (-\frac1{2})(x+1) + \frac{(-\frac1{2})((-\frac1{2})-1)}{2!}(x+1)^2 +$$\displaystyle \frac{(-\frac1{2})((-\frac1{2})-1)((-\frac1{2})-2)}{3!}(x+1)^3 + ... ]$

which is

$\displaystyle = x^2 [1 -\frac{x+1}{2} + \frac{3(x+1)^2}{{2^2}2!} - \frac{15(x+1)^2}{{2^3}3!} +$

$\displaystyle ... + \frac{(-1)^n(1\cdot3\cdot5\cdot\cdot\cdot(2n-1))(x+1)^n}{{2^n}n!}]$

I thought this might be okay until I checked my answer on WolframAlpha and got a different answer:

(x^2)*(2+x)^(-1/2) - Wolfram|Alpha
$\displaystyle \sum_{n=2}^{\infty}x^n 2^{3/2-n} \binom{-1/2}{-2+n}$

$\displaystyle \frac{x^2}{\sqrt{2}}-\frac{x^3}{4\sqrt{2}} + \frac{3x^4}{32\sqrt{2}}-\frac{5x^5}{128\sqrt{2}}+ ...$

So are both answers somehow correct? Or did I do something wrong?

Thanks,

Brian