JEE Mains · Maths · STD 11 - 7. binomial theoram
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
- A \(2041\)
- B \(2024\)
- C \(2014\)
- D \(2043\)
Answer & Solution
Correct Answer
(A) \(2041\)
Step-by-step Solution
Detailed explanation
\( \sum_{\mathrm{r}=1}^9 \frac{{ }^{11} \mathrm{C}_{\mathrm{r}}}{\mathrm{r}+1} \) \( =\frac{1}{12} \sum_{\mathrm{r}=1}^9{ }^{12} \mathrm{C}_{\mathrm{r}+1}\) \( =\frac{1}{12}\left[2^{12}-26\right]=\frac{2035}{6} \) \(\therefore \mathrm{m}+\mathrm{n}=2041\)
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