Binomial Theorem Question 6
Question 6 - 2024 (29 Jan Shift 1)
If $\frac{{ }^{11} C _1}{2}+\frac{{ }^{11} C _2}{3}+\ldots+\frac{{ }^{11} C _9}{10}=\frac{n}{m}$ with $\operatorname{gcd}(n, m)=1$, then $n+m$ is equal to
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Answer (2041)
Solution
$\sum _{r=1}^{9} \frac{{ }^{11} C _r}{r+1}$
$=\frac{1}{12} \sum _{r=1}^{9}{ }^{12} C _{r+1}$
$=\frac{1}{12}\left[2^{12}-26\right]=\frac{2035}{6}$
$\therefore m+n=2041$
