Question 25. If \frac{{}^{11}C_{1}}{2}+\frac{{}^{11}C_{2}}{3}+\cdots+\frac{{}^{11}C_{9}}{10}=\frac{n}{m} with \gcd(n,m)=1, then the value of n + m is
Answer: 2041
Explanation:
The given sum can be written as
\sum_{r=1}^{9}\frac{{}^{11}C_{r}}{r+1}.
Using the identity
\frac{{}^{11}C_{r}}{r+1}=\frac{1}{12}\,{}^{12}C_{r+1},
we get \sum_{r=1}^{9}\frac{{}^{11}C_{r}}{r+1}=\frac{1}{12}\sum_{r=1}^{9}{}^{12}C_{r+1}.
So, =\frac{1}{12}\left({}^{12}C_{2}+{}^{12}C_{3}+\cdots+{}^{12}C_{10}\right).
Now, \sum_{k=0}^{12}{}^{12}C_{k}=2^{12}.
Hence,
{}^{12}C_{2}+{}^{12}C_{3}+\cdots+{}^{12}C_{10}=2^{12}-({}^{12}C_{0}+{}^{12}C_{1}+{}^{12}C_{11}+{}^{12}C_{12}).
That is, =2^{12}-26.
Therefore,
\sum_{r=1}^{9}\frac{{}^{11}C_{r}}{r+1}=\frac{1}{12}(2^{12}-26)=\frac{2035}{6}.
So, n=2035,\ m=6.
Hence, n+m=2041.