JEE Main 2024 · MathsNumerical answerNumerical valueHardMulti-step

JEE Main 4 April 2024, Shift 1, Maths Q23

Question 23 of 90 in this shift, Maths question 23 of 30, Section B.

Let

a=1+\frac{{}^2C_2}{3!}+\frac{{}^3C_2}{4!}+\frac{{}^4C_2}{5!}+\ldots

b=1+\frac{{}^1C_0+{}^1C_1}{1!}+\frac{{}^2C_0+{}^2C_1+{}^2C_2}{2!}+\frac{{}^3C_0+{}^3C_1+{}^3C_2+{}^3C_3}{3!}+\ldots

. Then

\frac{2b}{a^2}

is equal to ________.

Official answer

NTA final key.