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7.Binomial Theorem
hard
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
$2043$
(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
