If $\sum_{r=0}^5\frac{{}^{11}C_{2r+1}}{2r+2}=\frac mn$, $\gcd(m,n)=1$, then $m-n$ is equal to _______

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $\frac{{}^{11}C_k}{k+1}=\frac{{}^{12}C_{k+1}}{12}$, so the sum is $\frac1{12}\sum_{j\ \mathrm{even}\ge2}{}^{12}C_j=\frac{2^{11}-1}{12}=\frac{2047}{12}$. Hence $m-n=2047-12=2035$. Answer: 2035.

Q. If $\sum_{r=0}^5\frac{{}^{11}C_{2r+1}}{2r+2}=\frac mn$, $\gcd(m,n)=1$, then $m-n$ is equal to _______

Explanation

Reviewed by Dr. Abubacker

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