If Σ_r=0^5 ^11 C_2r+1 2 r+2= m n, gcd( m, n)=1, then m- n is equal to

$\begin{aligned}
& (1+x)^{11}={ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+\cdots+{ }^{11} C_{11} x^{11} \
& \int_0^1(1+x)^{11} d x=\int_0^1\left({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+\cdots+{ }^{11} C_{11} x x^{11}\right) d x \
& \left.\left.\frac{(1-x)^{12}}{12}\right|0 ^1={ }^{11} C{0 x}+\frac{{ }^{11} C_1 x^2}{2}+\frac{{ }^{11} C_2 x}{3}+\cdots+\frac{{ }^{11} C_{11} x^{12}}{12}\right]0^1 \
& \frac{2^{12}-1}{12}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\cdots+\frac{C{11}}{12} \ldots(1)
\end{aligned}$Now,$Now,\begin{aligned}
& \int_{-1}^0(1+x)^{11} d x=\int_{-1}^0\left({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+\cdots+{ }^{11} C_{11} x^{11}\right) d x \
& \left.\left.\frac{(1+x)^{12}}{12}\right]{-1}^0={ }^{11} C_0 x+\frac{{ }^{11} C_1 x^2}{2}+\frac{{ }^{11} C_2 x^3}{3}+\cdots+\frac{{ }^{11} C{11} x^{12}}{12}\right]{-1}^0
\end{aligned}$$\frac{1}{12}=C_0-\frac{C_1}{2}+\frac{C_2}{3} \cdots$$\begin{aligned}
& (1)-(2) \
& =\frac{2^{12}-2}{12}=2\left[\frac{C_1}{2}+\frac{C_3}{4}+\cdots\right] \
& \Rightarrow \sum{r=0}^5 \frac{C_{2 r+1}}{2 r+2}=\frac{2^{11}-1}{12}=\frac{2047}{12}=\frac{m}{n} \
& =2047-12=2035
\end{aligned}$