Let a₁=b₁=1, aₙ=aₙ₋₁+2 and bₙ=aₙ+bₙ₋₁ for every
natural number n ≥ 2. Then Σₙ₌₁¹⁵ aₙ.bₙ is equal to ___________.

Solution

Maths sir · 4 steps

Find a formula for each sequence first: aₙ is the odd numbers and bₙ adds them up, so bₙ = n².

1. Find a_n. a₁ = 1 and each term adds 2, so aₙ is an AP: aₙ = 1 + 2(n-1) = 2n - 1. These are the odd numbers 1, 3, 5, …
2. Find b_n. bₙ = bₙ₋₁ + aₙ with b₁ = a₁ = 1, so bₙ = a₁ + a₂ + … + aₙ. The sum of the first n odd numbers is n², so bₙ = n².

   Σₙ₌₁¹⁵ aₙ bₙ = Σₙ₌₁¹⁵ (2n-1)n² = 2Σₙ₌₁¹⁵ n³ - Σₙ₌₁¹⁵ n²

3. Put in numbers. Σ n³ = ((15 × 16)/2)² = 120² = 14400 and Σ n² = (15 × 16 × 31)/6 = 1240.
4. So. The sum is 2 × 14400 - 1240 = 28800 - 1240 = 27560.

The common slip: 13160. Dropping the factor 2 in (2n-1)n² gives 14400 - 1240 = 13160, not 27560.

The idea to keep. When one sequence is built by adding up another, the second is just the partial sum of the first. Get both closed forms, then split the product into standard sums of powers.

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