If $\sum_{r=0}^{5} \frac{^{11}C_{2r+1}}{2r+2} = \frac{m}{n}$, $\gcd(m,n)=1$, then $m-n$ is equal to:
Solution:
Given sum:
$$ S = \sum_{r=0}^{5} \frac{^{11}C_{2r+1}}{2r+2} $$
Using the property: $\frac{^nC_k}{k+1} = \frac{^{n+1}C_{k+1}}{n+1}$
Here $n=11$ and $k=2r+1$. So, $k+1 = 2r+2$.
$$ \therefore \frac{^{11}C_{2r+1}}{2r+2} = \frac{^{12}C_{2r+2}}{12} $$
Substituting this in the summation:
$$ S = \sum_{r=0}^{5} \frac{^{12}C_{2r+2}}{12} $$
$$ S = \frac{1}{12} \left[ ^{12}C_2 + ^{12}C_4 + ^{12}C_6 + \dots + ^{12}C_{12} \right] $$
We know that sum of even binomial coefficients is:
$$ ^{12}C_0 + ^{12}C_2 + ^{12}C_4 + \dots + ^{12}C_{12} = 2^{12-1} = 2^{11} $$
So, the term inside the bracket is:
$$ (^{12}C_2 + \dots + ^{12}C_{12}) = 2^{11} – ^{12}C_0 $$
$$ = 2^{11} – 1 \quad (\because ^{12}C_0 = 1) $$
Now putting this value back in $S$:
$$ S = \frac{1}{12} (2^{11} – 1) $$
$$ S = \frac{2048 – 1}{12} = \frac{2047}{12} $$
Comparing with $\frac{m}{n}$, we get $m = 2047$ and $n = 12$.
$$ \therefore m – n = 2047 – 12 = 2035 $$
Ans. (2035)