JEE Main20244 Apr 2024Morning ShiftMathematicsBinomial TheoremActual
Let a=1+ ^2 C ₂ 3 ! + ^3 C ₂ 4 ! + ^4 C ₂ 5 ! + , b =1+ ^1 C ₀+ ^1 C ₁ 1 ! + ^2 C ₀+ ^2 C ₁+ ^2 C ₂ 2 ! + ^3 C ₀+ ^3 C ₁+ ^3 C ₂+ ^3 C ₃ 3 ! + Then 2 b a^2 is equal to
Correct answer
0
Step-by-step solution
aligned & f ( x )=1+ (1+ x ) 1 ! + (1+ x )^2 2 ! + (1+ x )^3 3 ! + . . & e ^ (1+ x ) 1+ x = 1 1+ x +1+ (1+ x ) 2 ! + (1+ x )^2 3 ! + (1+ x )^2 4 ! & coef x ^2 in RHS : 1+ ^2 C ₂ 3 + ^3 C ₂ 4 + = a aligned coeff. x^2 in L.H.S. e (1+x+ x^2 2 ! ) (1-x+ x^2 2 ! ) is e - e + e 2 ! = a aligned & b =1+ 2 1 ! + 2^2 2 ! + 2^3 3 ! + = e ^2 & 2 ~b a ^2 =8 aligned