If $\displaystyle\sum_{r=0}^{5}\dfrac{\binom{11}{2r+1}}{2r+2}=\dfrac{m}{n},\ \gcd(m,n)=1,$ then $m-n$ is equal to \rule{2cm}{0.4pt}.
Step-by-Step Solution
Key Concept: $\int_{0}^{1}(1+x)^{11}\,dx=\sum_{k=0}^{11}\dfrac{\binom{11}{k}}{k+1}.$ The integral over $[-1,0]$ flips signs on odd $k$. Subtract to isolate the odd-$k$ part.
$\displaystyle\int_{0}^{1}(1+x)^{11}\,dx=\frac{2^{12}-1}{12}=\frac{4095}{12}=\sum_{k=0}^{11}\dfrac{\binom{11}{k}}{k+1}.$
$\displaystyle\int_{-1}^{0}(1+x)^{11}\,dx=\frac{1}{12}=\sum_{k=0}^{11}\dfrac{(-1)^{k}\binom{11}{k}}{k+1}.$
Subtract: isolates odd $k$:
$$\frac{4095-1}{12}=\frac{4094}{12}=2\sum_{r=0}^{5}\dfrac{\binom{11}{2r+1}}{2r+2}\Rightarrow \sum_{r=0}^{5}\dfrac{\binom{11}{2r+1}}{2r+2}=\dfrac{2047}{12}.$$
$\gcd(2047,12)=1$, so $m=2047,\,n=12,\,m-n=2035.$
Correct Answer: 2035