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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

  1. A \(2041\)
  2. B \(2024\)
  3. C \(2014\)
  4. D \(2043\)
Verified Solution

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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