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7.Binomial Theorem
hard
यदि $\frac{{ }^{11} \mathrm{C}_1}{2}+\frac{{ }^{11} \mathrm{C}_2}{3}+\ldots . .+\frac{{ }^{11} \mathrm{C}_9}{10}=\frac{\mathrm{n}}{\mathrm{m}}$ है तथा $\operatorname{gcd}(\mathrm{n}, \mathrm{m})=1$ है, तो $\mathrm{n}+\mathrm{m}$ बराबर है ............
A
$2041$
B
$2024$
C
$2014$
D
$2043$
(JEE MAIN-2024)
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} $
$\therefore \mathrm{m}+\mathrm{n}=2041$
Standard 11
Mathematics
