What is the ratio of the $(r+1)$st term to the $r$th term in the expansion of $\left(1+\frac{x}{2}\right)^n$?

Question

What is the ratio of the $(r+1)$st term to the $r$th term in the expansion of $\left(1+\frac{x}{2}\right)^n$?

It seemed pretty easy and I soon came to the answer $\dfrac{x(n-r)}{2(r+1)}$

However this is wrong, any help would be appreciated

Answer 1

By the binomial theorem, $$ \left(1+\frac x2\right)^n = \sum_{k=0}^n \binom nk \left(\frac x2\right)^k. $$ So for $1\leqslant r\leqslant n-1$ we have the ratio of the coefficient of $\frac x2^{r+1}$ over the coefficienct of $\frac x2^r$ as $$ \binom n{r+1}/\binom nr = \frac{n!}{(r+1)!(n-r-1)!} \cdot \frac{r!(n-r)!}{n!} = \frac{n-r}{r+1}. $$