Let α=1²+4²+8²+13²+19²+26²+… upto 10 terms and β=Σn=1¹⁰ n⁴. If 4 α-β=55 k+40, then k is equal to __________.

Solution

Maths sir · 4 steps

The bases 1, 4, 8, 13, … have differences 3, 4, 5, …, so find their nth term and square it.

1. General term. The differences of 1, 4, 8, 13, 19, 26 are 3, 4, 5, 6, 7, so the bases grow quadratically. Fitting tₙ = an² + bn + c to the first three terms gives tₙ = (n² + 3n - 2)/2. Check: t₄ = (16 + 12 - 2)/2 = 13.
2. Square and expand. Each term of α is tₙ² = (n²+3n-2)²/4, so 4α = Σ (n²+3n-2)². The n⁴ part cancels against β.

   4α - β = Σₙ₌₁¹⁰(6n³ + 5n² - 12n + 4)

3. Use standard sums. For n = 1 to 10: Σ n³ = 3025, Σ n² = 385, Σ n = 55, and the constant adds 4 × 10 = 40. So 4α - β = 18150 + 1925 - 660 + 40 = 19415 + 40.
4. So. 19415 = 55 × 353, so 4α - β = 55 × 353 + 40 and k = 353.

The idea to keep. When the differences of a sequence form an AP, its nth term is a quadratic in n. Find it, expand, and use Σ n, Σ n², Σ n³.

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