Think this is a really easy question except I can't see a way to answer this.
Would you consider coefficients like: $$\binom {n}{r-2},\binom {n}{r-1}, \binom {n}{r}, \binom {n}{r+1} $$
However, there is no total sum that any random 4 consecutive coefficients would add up to, so I'm not sure.
Hint. Note that $$\frac{\binom{n}{r}}{\binom{n}{r}+\binom{n}{r+1}}=\frac{1}{1+\frac{n!}{(n-r-1)!(r+1)!}\cdot \frac{(n-r)!(r)!}{n!}}=\frac{1}{1+\frac{n-r}{r+1}}=\frac{r+1}{n+1}.$$
P.S. You can also use the fact (see Matthew Leingang's comment) that $$\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}=\frac{n+1}{r+1}\binom{n}{r}.$$
