JEE Main2018MathematicsBinomial TheoremActual
The coefficient of x 2 in the expansion of the product 2 - x 2 1 + 2 x + 3 x 2 6 + 1 - 4 x 2 6 is
Options
- A107
- B108
- C155
- D106
Correct answer
D. 106
Step-by-step solution
We have, ( 1 + 2 x + 3 x 2 ) 6 = ∑ r = 0 6   6 C r 2 x + 3 x 2 r = 6 C 0 + 6 C 1 2 x + 3 x 2 + 6 C 2 2 x + 3 x 2 2 +   … + 6 C 6 2 x + 3 x 2 6 ∴ Coefficient of x 2 = 18 + 60 = 78 Again, coefficient of x 2 in 1 - 4 x 2 6 = - 24 Constant term in 1 + 2 x + 3 x 2 6 + 1 - 4 x 2 6 is 1 + 1 = 2 . Thus, the required coefficient of x 2 = 2 (coefficient of x 2 ) in 1 + 2 x + 3 x 2 6 + 1 - 4 x 2 6 - constant term in 1 + 2 x + 3 x 2 6 + 1 - 4 x 2 6 = 2 78 - 24 - 2 = 106 .