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Find a if the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are equal
Solution:
It is known that (r + 1)th term, (Tᵣ ₊ ₁) in the binomial expansion of (a + b)n is given by Tᵣ ₊ ₁ = ⁿCᵣ aⁿ ⁻ ʳ bʳ.
In the expansion of (3 + ax)9,
Tᵣ ₊ ₁ = ⁹Cᵣ (3)⁹ ⁻ ʳ (ax)ʳ
= ⁹Cᵣ (3)⁹ ⁻ ʳ aʳ xʳ
Using this, the coefficients of x2 and x3 are ⁹C₂ (3)⁹ ⁻ ² a² and ⁹C₃ (3)⁹ ⁻ ³ a³ respectively.
It is given that these coefficients are equal. So
⁹C₂ (3)⁹ ⁻ ² a² = ⁹C₃ (3)⁹ ⁻ ³ a³
Using nCr formula,
9! / (2! 7!) 37 a2 = 9!/ (3! 6!) 36 a3
9! / (2! 7!) × (3! 6!) / 9! = (36 a3) / (37 a2)
3/7 = a/3
a = 9/7
NCERT Solutions Class 11 Maths Chapter 8 Exercise ME Question 2
Find a if the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are equal
Summary:
The value of a if the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are equal is 9/7
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