Let the coefficients of three consecutive terms in the binomial expansion of (1+2x)ⁿ be in the ratio 2 : 5 : 8. Then the coefficient of the term, which is in the middle of those three terms, is __________.

Solution

Maths sir · 5 steps

Divide consecutive coefficients: each ratio gives one equation, and two equations pin down n and r.

1. Set up the terms. In (1+2x)ⁿ the coefficient of xʳ is C(n, r)2ʳ. Take the three terms as the coefficients of xʳ⁻¹, xʳ and xʳ⁺¹, so the middle one is C(n, r)2ʳ.

   C(n, r)2ʳ/C(n, r-1)2ʳ⁻¹ = 2(n-r+1)/r

2. First ratio. Second to first is 5 : 2, so 2(n-r+1)/r = 5/2. This gives 4n + 4 = 9r.
3. Second ratio. Third to second is 8 : 5, so 2(n-r)/(r+1) = 8/5. This gives 5n = 9r + 4. Put 9r = 4n + 4 into it: 5n = 4n + 8, so n = 8 and r = 4.
4. Check. The three coefficients are C(8, 3)2³ = 448, C(8, 4)2⁴ = 1120 and C(8, 5)2⁵ = 1792. Dividing each by 224 gives 2 : 5 : 8, as required.
5. So. The middle coefficient is C(8, 4) × 2⁴ = 70 × 16 = 1120.

The common slip: 70. Reporting only the binomial coefficient C(8, 4) and forgetting the factor 2⁴ gives 70, not 1120.

The idea to keep. For consecutive terms, always take the ratio of neighbouring coefficients. The factorials cancel to a simple fraction in n and r, and the powers of the constant leave just one factor behind.

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