Find the $r^{th}$ term from the beginning and the $r^{th}$ term from the end of $(a+2x)^n$
$(a+2x)^n=[a(1+\dfrac{2x}{a})]^n=a^n(1+\dfrac{2x}{a})^n$
$(1+\dfrac{2x}{a})^n=1+^nC_1(\dfrac{2x}{a})+^nC_2(\dfrac{2x}{a})^2...+^nC_r(\dfrac{2x}{a})^r...$
So the $r^{th}$ term from the beginning is $a^n(^nC_r)(\dfrac{2x}{a})^r$. But the answer is given as $\dfrac{n(n-1)...(n-r+2)}{(r-1)!}a^{n-r+1}(2x)^{r-1}$.
As to the $r^{th}$ term from the end, I suppose I have to find a way to transform it so I can count it from the beginning? How would I go about doing that? Thanks.
We can expand $(a+2x)^n$ to get $${}^nC_0a^n(2x)^0+{}^nC_1a^{n-1}(2x)^1+{}^nC_2a^{n-2}(2x)^2+\ldots+{}^nC_na^0(2x)^n$$
As you can see the $r^\mathrm{th}$ term will be $${}^nC_{r-1}a^{n-(r-1)}(2x)^{r-1}={}^nC_{r-1}a^{n-r+1}(2x)^{r-1}$$ Note that the mistake you made is that you took the ${}^nC_0a^n(2x)^0$ term to be the $0^\mathrm{th}$ term rather than the $1^\mathrm{st}$, which gave you the $r^\mathrm{th}$ term as $${}^nC_ra^{n-r}(2x)^r={}^nC_ra^n\left(\frac{2n}a\right)^r$$ Since ${}^nC_r=\frac{n!}{r!(n-r)!}$, this can be rewritten as $$\begin{align}\frac{n!}{(r-1)!(n-r+1)!}a^{n-r+1}(2x)^{r-1}&=\frac{n(n-1)\ldots(n-r+2)(n-r+1)!}{(r-1)!(n-r+1)!}a^{n-r+1}(2x)^{r-1}\\&=\boxed{\frac{n(n-1)\ldots(n-r+2)}{(r-1)!}a^{n-r+1}(2x)^{r-1}}\end{align}$$ as required.