Binomial theorem (solve this please)

Question

If the two middle terms in the expansion of $( a+2b )^{2n+1}$ are equal then

(A) $\frac{a}{b}=\frac{1}{2}$ (B) $a=4b$ (C) $a=8b$ (d) $a=2b$

Answer 1

Short answer: with $n=0$, $(a+2b)^1=a+2b$, hence $a=2b$.

Answer 2

Binomial theorem: $(K + L)^m = sum {m \choose i}*K^iL^{m-i}$. Notice there are $m +1$ terms.

If there are $m= 2n + 1 $ there are $2n+2$ terms. The first $n+1$ terms are $i = 0..... n$ and the second $n+1$ terms are $n+1,...... 2n+1$. So the middle terms are $i =n$ and $i = n+1$.

So $(a+2b)^{2n+1} = sum {2n+1 \choose i}*a^{i}*(2b)^{2n+1 -i}$.

So for $i = n$ the term is ${2n+1 \choose n}*a^n*(2b)^{n+1}$.

For $i = n+1$ the term is ${2n+1 \choose n+1}*a^{n+1}*(2b)^n$

So we want ${2n+1 \choose n}*a^n*(2b)^{n+1} = {2n+1 \choose n+1}*a^{n+1}*(2b)^n$

One obvious thing to note:

${2n+1 \choose n} = \frac{(2n+1)!}{(2n+1-n)!n!} = \frac{(2n+1)!}{(n+1)!n!}=\frac{(2n+1)!}{(n+1)!(2n+1 - (n-1))!}= {2n+1 \choose n + 1}$

So, can you take it from there?