Solution
Related Formula
Coefficient of (r+1)th term in (1+x)ⁿ is ⁿCᵣ ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁Core Logic
Let the four consecutive binomial coefficients be ⁿCᵣ, ⁿCᵣ₊₁, ⁿCᵣ₊₂, ⁿCᵣ₊₃. Given these correspond to 2 - p, p, 2 - α, α respectively.
From the properties of combinations:
ⁿCᵣ + ⁿCᵣ₊₁ = (2 - p) + p = 2 ⇒ ⁿ⁺¹Cᵣ₊₁ = 2 (1)Similarly:
ⁿCᵣ₊₂ + ⁿCᵣ₊₃ = (2 - α) + α = 2 ⇒ ⁿ⁺¹Cᵣ₊₃ = 2 (2)Step 1: Identifying the Inconsistency
From equations (1) and (2):
ⁿ⁺¹Cᵣ₊₁ = ⁿ⁺¹Cᵣ₊₃ = 2By combination properties, if ⁿCₓ = ⁿCy and x ≠ y, then x + y = n. So, (r+1) + (r+3) = n+1 ⇒ 2r + 4 = n+1 ⇒ n = 2r + 3.
Substitute n back into the equality:
2r+4Cᵣ₊₁ = 2The binomial coefficient must be ≥ 2. However, for any valid integer r ≥ 0, 2r+4Cᵣ₊₁ grows very rapidly. Let's test small values: If r = 0, ⁴C₁ = 4 ≠ 2. If r = 1, ⁶C₂ = 15 ≠ 2. Hence, no integer values satisfy this condition, making the given data inherently inconsistent.
Step 2: Conclusion
Due to inconsistent data resulting in a mathematically impossible scenario, this question was treated as a Bonus/Dropped question.
Pattern Recognition
Adding consecutive binomial coefficients yields the sum from Pascal's triangle identity ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁. If identical sums yield small integer invariants like 2, they typically break bounding limits for ⁿCk combinations.
Chapter Mix
Class 11 Maths: Binomial Theorem