BITSAT2019MathematicsBinomial TheoremActual
If the coefficient of x 3 and x 4 in the expansion of 1 + a x + b x 2 1 - 2 x 18 in powers of x are both zero, then a , b is equal to
Options
- A16 , 251 3
- B14 , 251 3
- C14 , 272 3
- D16 , 272 3
Correct answer
D. 16 , 272 3
Step-by-step solution
We have, 1 + a x + b x 2 1 - 2 x 18 1 + a x + b x 2 1 - C 1 18 2 x + C 2 18 2 x 2 - C 3 18 2 x 3 + C 4 18 2 x 4 … Coefficient of x 3 is - C 3 18 2 3 + a · C 2 18 2 2 - b 18 C 1 2 and coefficient of x 4 is - C 4 18 2 4 - C 3 18 2 3 a + C 2 18 2 2 b Coefficient of x 3 and x 4 are zero. ∴   - C 3 18 2 3 + C 2 18 2 2 a - C 1 18 2 b = 0 ⇒   4 × 17 × 16 3 × 2 - 17 a + b = 0 . . . i and 80 - 32 3 a + b = 0 . . . ii Solving Eqs. ( i ) and ( ii ) , we get a = 16 , b = 272 3