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