JEE Main 20244 Apr, Morning shiftMathematicsHardBinomial Theorem and Its Simple Applications131× since 2002Let a=1+2C23!+3C24!+4C25!+....,b=1+1C0+1C11!+2C0+2C1+2C22!+3C0+3C1+3C2+3C33!+....a=1+\frac{{ }^2 \mathrm{C}_2}{3 !}+\frac{{ }^3 \mathrm{C}_2}{4 !}+\frac{{ }^4 \mathrm{C}_2}{5 !}+...., \mathrm{b}=1+\frac{{ }^1 \mathrm{C}_0+{ }^1 \mathrm{C}_1}{1 !}+\frac{{ }^2 \mathrm{C}_0+{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2}{2 !}+\frac{{ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3}{3 !}+....a=1+3!2C2+4!3C2+5!4C2+....,b=1+1!1C0+1C1+2!2C0+2C1+2C2+3!3C0+3C1+3C2+3C3+.... Then 2ba2\frac{2 b}{a^2}a22b is equal to _________.Your answerType your answer. Integers and decimals are accepted.CheckSkip · show solution