Let f(x)=Σk=1¹⁰ k xᵏ, x ∈ ℝ. If 2 f(2)+f′(2)=119(2)ⁿ+1 then n is equal to ___________

Solution

Maths sir · 4 steps

Merge both sums into one term, (k²+4k) 2ᵏ⁻¹, then make it telescope.

1. Write both sums. f(2) = Σ k 2ᵏ, so 2f(2) = Σ 4k 2ᵏ⁻¹. Differentiating term by term gives f'(x) = Σ k² xᵏ⁻¹, so f'(2) = Σ k² 2ᵏ⁻¹. Adding them, the general term is (k²+4k) 2ᵏ⁻¹ for k = 1 to 10.
2. Find a telescoping form. Try g(k) = (k²+2k-1) 2ᵏ. Then g(k-1) = (k²-2) 2ᵏ⁻¹, and subtracting gives exactly our term.

   g(k)-g(k-1) = 2ᵏ⁻¹[2(k²+2k-1)-(k²-2)] = (k²+4k) 2ᵏ⁻¹

3. Add up. The sum from k=1 to 10 collapses to g(10)-g(0). Here g(10) = (100+20-1) 2¹⁰ = 119 × 2¹⁰ and g(0) = -1, so 2f(2)+f'(2) = 119 × 2¹⁰ + 1 = 121857.
4. So. Comparing with 119 × 2ⁿ + 1 gives n = 10.

The idea to keep. For a sum of (polynomial in k) × rᵏ, guess a closed form of the same type, (polynomial) × rᵏ, and match coefficients so the terms telescope.

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