In the expansion of $(3+4x)^n$ the coefficients of $x^2$ and $x^3$ are in the ratio of $3:4$. Find the value of $n$ Every time I do this question, I get $5$ as my answer, whilst the  answers says $4$? Can someone please check this:
"In the expansion of $(3+4x)^n$ the coefficients of $x^2$ and $x^3$ are in the ratio of $3:4$. Find the value of $n$."

In fact in the general case, if you take the ratio of the $k^{th}$ and $(k+1)^{th}$ terms as being 3:4, then this leads to:
$$\begin{align}
\frac{{n\choose k}\times3^{n-k}\times4^{k}}{{n\choose k+1}\times3^{n-k-1}\times4^{k+1}}&=\frac{3}{4}\\
\therefore \frac{{n\choose k}\times3^{n-k}\times4^{k+1}}{{n\choose k+1}\times3^{n-k}\times4^{k+1}}&=1\\
\therefore {n\choose k}&={n\choose k+1}\\
\therefore \frac{n!}{k!(n-k)!}&=\frac{n!}{(k+1)!(n-k-1)!}\\
\therefore n-k&=k+1\\
\therefore n&=2k+1
\end{align}$$
Which means $n$ has to be an odd number, so $n=4$ cannot be correct.
