JEE Main 2024 — Binomial Theorem Question with Solution

Given,

$\frac{{}^{11}C_1}{2}+\frac{{\displaystyle {}^{11}C_2}}{{\displaystyle 3}}+\frac{{\displaystyle {}^{11}C_3}}{{\displaystyle 4}}.........+\frac{{\displaystyle {}^{11}C_9}}{{\displaystyle 10}}=\frac{{\displaystyle m}}{{\displaystyle n}}$

Now, we know that,

${\left(1+x\right)}^{11}={}^{11}C_0+{}^{11}C_{1}x+{}^{11}C_2{x}^2+..........{}^{11}C_{11}{x}^{11}$

Now integrating the above expression both side we get,

${\int }_0^1{\left(1+x\right)}^{11}dx={\int }_0^1{}^{11}C_0+{}^{11}C_{1}x+{}^{11}C_2{x}^2+..........{}^{11}C_{11}{x}^{11}dx$

$\Rightarrow {\left[\frac{{\left(1+x\right)}^{12}}{12}\right]}_0^1={\left[{}^{11}C_{0}x+{}^{11}C_1\frac{x^2}{2}+{}^{11}C_2\frac{x^3}{3}+..........{}^{11}C_{11}\frac{x^{12}}{12}\right]}_0^1$

$\Rightarrow \frac{2^{12}-1}{12}=1+\frac{{}^{11}C_1}{2}+\frac{{\displaystyle {}^{11}C_2}}{{\displaystyle 3}}+\frac{{\displaystyle {}^{11}C_3}}{{\displaystyle 4}}.........+\frac{{\displaystyle {}^{11}C_9}}{{\displaystyle 10}}+\frac{11}{11}+\frac{1}{12}$​​​​​​

$\Rightarrow \frac{2^{12}-1}{12}-1-1-\frac{1}{12}=\frac{{}^{11}C_1}{2}+\frac{{\displaystyle {}^{11}C_2}}{{\displaystyle 3}}+\frac{{\displaystyle {}^{11}C_3}}{{\displaystyle 4}}.........+\frac{{\displaystyle {}^{11}C_9}}{{\displaystyle 10}}$

$\Rightarrow \frac{{}^{11}C_1}{2}+\frac{{\displaystyle {}^{11}C_2}}{{\displaystyle 3}}+\frac{{\displaystyle {}^{11}C_3}}{{\displaystyle 4}}.........+\frac{{\displaystyle {}^{11}C_9}}{{\displaystyle 10}}=\frac{2^{12}-2-24}{12}$

$\Rightarrow \frac{{}^{11}C_1}{2}+\frac{{\displaystyle {}^{11}C_2}}{{\displaystyle 3}}+\frac{{\displaystyle {}^{11}C_3}}{{\displaystyle 4}}.........+\frac{{\displaystyle {}^{11}C_9}}{{\displaystyle 10}}=\frac{4096-26}{12}$

$\Rightarrow \frac{{}^{11}C_1}{2}+\frac{{\displaystyle {}^{11}C_2}}{{\displaystyle 3}}+\frac{{\displaystyle {}^{11}C_3}}{{\displaystyle 4}}.........+\frac{{\displaystyle {}^{11}C_9}}{{\displaystyle 10}}=\frac{4070}{12}$

$\Rightarrow \frac{{}^{11}C_1}{2}+\frac{{\displaystyle {}^{11}C_2}}{{\displaystyle 3}}+\frac{{\displaystyle {}^{11}C_3}}{{\displaystyle 4}}.........+\frac{{\displaystyle {}^{11}C_9}}{{\displaystyle 10}}=\frac{2035}{6}$

Hence, on comparing we get, $m+n=2041$