JEE MAIN - Mathematics (2024 - 29th January Morning Shift - No. 25)
$$\text { If } \frac{{ }^{11} C_1}{2}+\frac{{ }^{11} C_2}{3}+\ldots+\frac{{ }^{11} C_9}{10}=\frac{n}{m} \text { with } \operatorname{gcd}(n, m)=1 \text {, then } n+m \text { is equal to }$$ _______.
Answer
2041
Explanation
$$\begin{aligned}
& \sum_{\mathrm{r}=1}^9 \frac{{ }^{11} \mathrm{C}_{\mathrm{r}}}{\mathrm{r}+1} \\
& =\frac{1}{12} \sum_{\mathrm{r}=1}^9{ }^{12} \mathrm{C}_{\mathrm{r}+1} \\
& =\frac{1}{12}\left[2^{12}-26\right]=\frac{2035}{6} \\
& \therefore \mathrm{m}+\mathrm{n}=2041
\end{aligned}$$
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