Finding the $N^{th}$ term In binomial expansion for arbitrary power

Question

How do I find the $n^{th}$ term in the binomial expansion of any index.

Ex: Find the $4^{th}$ term of $(1+2x)^{-1/2}$.

Answer 1

Hint:

Start from the general expansion of $`\,(1+u)^\alpha$: $$(1+u)^\alpha=1+\alpha u+\frac{\alpha(\alpha-1)}{2!} u^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!} u^3+\dots+\frac{\alpha(\alpha-1)\dotsm(\alpha-n+1)}{n!} u^n+\dotsm$$

If you substitute $\,\alpha=-\dfrac 12$ and $u=2x$, you'll ind the coefficient of $x^n$ is eventually the signed product of the first $n$ odd integers: $$(-1)^n1\cdot 3\cdot 5\dotsm(2n-1).$$ This results from the simplification of: $$\binom{-\frac 12}{n} (2x)^n $$ where $\dbinom{-\frac 12}{n}$ is the *generalised binomial coefficient: $$\dbinom{-\frac 12}{n}=\frac{-\frac12\bigl(-\frac12-1\bigr)\bigl(-\frac12-2\bigr)\dotsm\bigl(-\frac12-n+1\bigr)}{n!}$$