Question

If the first three terms in the expansion of (a + b)ⁿ are 27, 54, and 36 respectively, then find a, b and n.

Answer 1

Since, 

(a + b)_n = ⁿC₀ aⁿ + ⁿC1 a^{n – 1} b + ⁿC₂ a^{n – 2} b ² +…+ ⁿCn bⁿ

Given that, 

T1 =ⁿC₀ a ⁿ = 27 

⇒ aⁿ= 27   (i)

T₂ = ⁿC₁ a^{n – 1} b = 54 

⇒ na^{n – 1} b = 54 

(ii) T₃ = ⁿC₂ a ^{n – 2} b ² = 36

\(\frac{n(n-1)}{2}\) a^n-2 b² =36

Dividing (ii) by (i), we get

\(\frac{na^{n-1}b}{a_n}\) = \(\frac{54}{27}\)

\(\frac{nb}{a}\) = 2

Dividing (iii) by (ii), we get

⇒ \(\frac{b}{a}=\frac{2}{3}\)

From (iv) and (v), we get

 n = 3 (vi) From (i) and (vi), we get 

a = 3 (vii) From (v) and (vii), we get

 b = 2