Let the coefficients of third, fourth and fifth terms in the expansion of ( x + a/x² )ⁿ,x ≠ 0, be in the ratio 12 : 8 : 3. Then the term independent of x in the expansion, is equal to ___________. (Round off to the nearest integer)

Solution

Maths sir · 4 steps

Divide neighbouring coefficients: the ratios give two easy equations in n and a.

1. General term. Tᵣ₊₁ = C(n, r) xⁿ⁻ʳ(a/x²)ʳ = C(n, r) aʳ xⁿ⁻³ʳ. So the 3rd, 4th and 5th coefficients are C(n, 2)a², C(n, 3)a³ and C(n, 4)a⁴.

   C(n, 2)a²/C(n, 3)a³ = 3/(n-2)a = 12/8, C(n, 3)a³/C(n, 4)a⁴ = 4/(n-3)a = 8/3

2. Find n and a. These give (n-2)a = 2 and (n-3)a = 3/2. Subtracting the second from the first gives a = 1/2. Then n - 2 = 4, so n = 6.
3. Term free of x. The power of x is zero when n - 3r = 0, so 6 - 3r = 0 and r = 2. That term is C(6, 2)(1/2)² = 15 × 1/4 = 3.75.
4. So. The term independent of x is 15/4 = 3.75, which rounds to 4.

The common slip: 15. Forgetting the factor a² = 1/4 in the constant term gives C(6, 2) = 15, not 3.75.

The idea to keep. When coefficients are given as a ratio, divide consecutive ones. The ratio C(n, r)/C(n, r+1) = (r+1)/(n-r) turns the factorials into simple linear equations.

Answer from the NTA final answer key. Solution by PrepWiser.