If frac 11 C 1 2 +frac 11 C 2 3 +ldots . .+frac 11 C 9 10 =frac n m with gcd (n, m)=1 , then n+m is equal to

Mathematics · Binomial Theorem

JEE Main 2024 — 29 January, Shift 1 — Question 28

If $\frac{{ }^{11} \mathrm{C}_{1}}{2}+\frac{{ }^{11} \mathrm{C}_{2}}{3}+\ldots . .+\frac{{ }^{11} \mathrm{C}_{9}}{10}=\frac{n}{\mathrm{~m}}$ with $\operatorname{gcd}(n, m)=1$ , then $n+m$ is equal to

Answer: 2041

Numerical answer — enter this value.

Step-by-step solution

$\sum_{\mathrm{r}=1}^{9} \frac{{ }^{11} \mathrm{C}_{\mathrm{r}}}{\mathrm{r}+1}$

$=\frac{1}{12} \sum_{\mathrm{r}=1}^{9}{ }^{12} \mathrm{C}_{\mathrm{r}+1}$

$=\frac{1}{12}\left[2^{12}-26\right]=\frac{2035}{6}$

$\therefore \mathrm{m}+\mathrm{n}=2041$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Series involving sum & product of Binomial Coefficients