Step 1: Write the sum neatly.
The sum is $S=\displaystyle\sum_{r=0}^{n}(r+1)\dfrac{\binom{n}{r}}{2^n}.$ Pull out $\dfrac1{2^n}$ and split $(r+1)$ into $r$ and $1$.
Step 2: Split into two known sums.
$S=\dfrac1{2^n}\left(\displaystyle\sum_{r=0}^{n} r\binom{n}{r}+\sum_{r=0}^{n}\binom{n}{r}\right).$
Step 3: Use standard identities.
We know $\displaystyle\sum_{r=0}^{n}\binom{n}{r}=2^n$ and $\displaystyle\sum_{r=0}^{n} r\binom{n}{r}=n\,2^{n-1}.$
Step 4: Put them in.
$S=\dfrac{n\,2^{n-1}+2^n}{2^n}.$ Split the fraction: $\dfrac{n\,2^{n-1}}{2^n}+\dfrac{2^n}{2^n}=\dfrac{n}{2}+1.$
Step 5: Use the given value.
We are told $S=16$, so $\dfrac{n}{2}+1=16$, giving $\dfrac{n}{2}=15.$
Step 6: Solve.
$n=30.$ \[ \boxed{30} \]