Let the coefficient of xʳ in the expansion of (x+3)ⁿ⁻¹+(x+3)ⁿ⁻²(x+2)+(x+3)ⁿ⁻³(x+2)²+… … … .+(x+2)ⁿ⁻¹ be αᵣ. If Σr=0ⁿ αᵣ=βⁿ-γⁿ, β, γ ∈ ℕ, then the value of β²+γ² equals _________.

Solution

Maths sir · 4 steps

This is a geometric series in disguise: the ratio of terms is (x+2)/(x+3), so it collapses to one line.

1. Spot the pattern. Put a = x+3 and b = x+2. The sum is aⁿ⁻¹ + aⁿ⁻²b + … + bⁿ⁻¹, which has the same shape as the factor in aⁿ - bⁿ = (a-b)(aⁿ⁻¹ + aⁿ⁻²b + … + bⁿ⁻¹).

   Σₖ₌₀ⁿ⁻¹ aⁿ⁻¹⁻ᵏbᵏ = (aⁿ - bⁿ)/(a-b) = (x+3)ⁿ - (x+2)ⁿ

2. Sum of coefficients. Here a - b = 1, so the whole expression is just (x+3)ⁿ - (x+2)ⁿ. The sum of all its coefficients Σ αᵣ is its value at x = 1, which is 4ⁿ - 3ⁿ.
3. Match the form. Comparing with βⁿ - γⁿ gives β = 4 and γ = 3. The coefficient αₙ is 0 because the xⁿ terms cancel, so it does not change the sum.
4. So. β² + γ² = 16 + 9 = 25.

The common slip: 13. Putting x = 0 instead of x = 1 gives 3ⁿ - 2ⁿ, so β = 3, γ = 2 and the answer 13. Setting x = 0 gives only the constant term, not the sum of all coefficients.

The idea to keep. When a sum steps down powers of one bracket and steps up powers of another, think (aⁿ - bⁿ)/(a-b). For the sum of all coefficients, put x = 1.

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