Step 1: Look at each term.
The terms are $\dfrac34,\dfrac{15}{16},\dfrac{63}{64},\dots$ Notice each one is just less than $1$. Write them as $1$ minus a small fraction.
Step 2: Spot the pattern.
We see \[ \tfrac34=1-\tfrac14,\quad \tfrac{15}{16}=1-\tfrac1{16},\quad \tfrac{63}{64}=1-\tfrac1{64}. \] The small parts are $\dfrac1{4},\dfrac1{16},\dfrac1{64}=\dfrac1{4^1},\dfrac1{4^2},\dfrac1{4^3}$.
Step 3: Split the sum.
So the $k$-th term is $1-\dfrac1{4^k}$. Adding $n$ terms, \[ S_n=\underbrace{(1+1+\dots)}_{n\text{ times}}-\left(\frac1{4}+\frac1{4^2}+\dots+\frac1{4^n}\right)=n-\sum_{k=1}^{n}\frac1{4^k}. \]
Step 4: Sum the small geometric part.
The bracket is a GP with first term $\dfrac14$ and ratio $\dfrac14$: \[ \sum_{k=1}^{n}\frac1{4^k}=\frac{\tfrac14\left(1-\tfrac1{4^n}\right)}{1-\tfrac14}=\frac13\left(1-\frac1{4^n}\right). \]
Step 5: Use the given total.
So $S_n=n-\dfrac13\left(1-\dfrac1{4^n}\right)=\dfrac{939}{256}$. Trying $n=4$: $4-\dfrac13\left(1-\dfrac1{256}\right)=4-\dfrac13\cdot\dfrac{255}{256}=4-\dfrac{85}{256}=\dfrac{939}{256}.$ It matches, so $n=4$.
Step 6: Get the asked value.
The question wants $5n$. \[ 5n=5\times 4=20. \] \[ \boxed{20} \]