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(1+2x) ^(5/2)
So far i have
1 + 5x – 15/2 x² but am not sure about the x³
thanks
The generalized binomial formula $\displaystyle (a+ b)^c$ where c does not have to be an integer says
$\displaystyle (a+ b)^c= \sum_{n=0}^\infty \left(\begin{array}{c}c \\ n\end{array}\right) x^n y^{c-n}$
where, with c not an integer, we interpret $\displaystyle \left(\begin{array}{c}c \\ n\end{array}\right)$ as $\displaystyle \frac{c(c-1)(c-2)\cdot\cdot\cdot(c-n+1)}{n!}$, for n= 0, 1 for n= 0.
In this case, with c= 5/2, $\displaystyle \left(\begin{array}{c}5/2 \\ 0 \end{array}\right)= 1$, so the $\displaystyle x^0$ term is 1, $\displaystyle \left(\begin{array}{c}5/2 \\ 1\end{array}\right)= 5/2$ so the "x" term is (5/2)(2x)= 5/2 x as you say. $\displaystyle \left(\begin{array}{c}5/2 \\ 2\end{array}\right)= ((5/2)(3/2)/2)(2x)^2= (15/2)x^2$ as you say. $\displaystyle \left(\begin{array}{c}5/2 \\ 3\end{array}\right)= ((5/2)(3/2)(1/2)/6)(4x)^3= [(15/8)/6](64x^3)= 20x^3$
Check my arithmetic!