Jee Main 2024 29 01 2024 Shift 1 - Question15

Answer: (2041)

Solution:

$(1+x)^{11}={ }^{11} C _0+{ }^{11} C _1 x+{ }^{11} C _2 x^{2}+\ldots .+{ }^{11} C _{11} x^{11}$

$$ \begin{aligned} & \int _0^{1}(1+x)^{11} d x={ }^{11} C _0 x+\frac{{ }^{11} C _1 x^{2}}{2}+\frac{{ }^{11} C _2 x^{3}}{3}+\ldots . \\ & \left.+\frac{{ }^{11} C _9 x^{10}}{10}+\frac{{ }^{11} C _{10} x^{11}}{11}+\frac{{ }^{11} C _{11} x^{12}}{12}\right] _0^{1} \\ & \left.\frac{(1+x)^{12}}{12}\right] _0^{1}={ }^{11} C _0+\frac{{ }^{11} C _1}{2}+\frac{{ }^{11} C _2}{3}+\ldots+\frac{{ }^{11} C _9}{10}+\frac{{ }^{11} C _{10}}{11}+\frac{{ }^{11} C _{11}}{12} \\ & \frac{2^{12}-1}{12}-1-1-\frac{1}{12}=\frac{{ }^{11} C _1}{2}+\frac{{ }^{11} C _2}{3}+\ldots+\frac{{ }^{11} C _{10}}{11} \\ & =\frac{2^{12}-2-24}{12} \\ & =\frac{2^{12}-26}{12}=\frac{4070}{12}=\frac{2035}{6}=\frac{m}{n} \\ & m+n=2035+6=2041 \end{aligned} $$