If the coefficients of x3 and x4 in the expansion of (1 + ax + bx2) (1 – 2x)18 in powers of x are both zero, then (a, b) is equal to

Engineering

Mathematics

Introduction to Binomial Theorem

Question

If the coefficients of x³ and x⁴ in the expansion of (1 + ax + bx²) (1 – 2x)¹⁸ in powers of x are both zero, then (a, b) is equal to

$(14, \frac{272}{3})$

$(14, \frac{251}{3})$

$(16, \frac{251}{3})$

$(16, \frac{272}{3})$

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coefficient of x³ in (1 + ax + bx²) (1 – 2x)¹⁸

= coefficient of x³ in (1 – 2x)¹⁸

+ a · coefficient of x² in (1 – 2x)¹⁸

+ b · coefficient of x in (1 – 2x)¹⁸ = 0

⇒ ¹⁸C₃ (– 2)³ + a · ¹⁸C₂ (– 2)²

+ b · ¹⁸C₁ (– 2)¹ = 0 ……. (1)

and

coefficient of x⁴ in (1 + ax + bx²) (1 – 2x)¹⁸

= coefficient of x⁴ in (1 – 2x)¹⁸

+ a · coefficient of x³ in (1 – 2x)¹⁸

+ b · coefficient of x² in (1 – 2x)¹⁸ = 0

⇒ ¹⁸C₄ (– 2)⁴ + a · ¹⁸C₃ (– 2)³

+ b · ¹⁸C₂ (– 2)² = 0 …….(2)

$∴$ From (1) and (2)

we get, a = 16, $b = \frac{272}{3}$ .