Let < aₙ > be a sequence such that a₁+a₂+…+aₙ=(n²+3 n)/(n+1)(n+2). If 28 Σk=1¹⁰ 1/aₖ=p₁ p₂ p₃ … pₘ, where p₁, p₂, … ., pₘ are the first m prime numbers, then m is equal to

Solution

Maths sir · 5 steps

Get aₙ from Sₙ - Sₙ₋₁, then 1/aₖ is a product of three consecutive integers that telescopes.

1. Tidy the sum. Since n² + 3n = (n+1)(n+2) - 2, the given sum is Sₙ = 1 - 2/(n+1)(n+2).
2. Find the term. aₙ = Sₙ - Sₙ₋₁ = 2/n(n+1) - 2/(n+1)(n+2) = 4/n(n+1)(n+2). This works for n = 1 too, since a₁ = 4/6 = 2/3 = S₁. So 1/aₖ = k(k+1)(k+2)/4.

   Σₖ₌₁ⁿ k(k+1)(k+2) = n(n+1)(n+2)(n+3)/4

3. Why it telescopes. Each term is k(k+1)(k+2) = 1/4[k(k+1)(k+2)(k+3) - (k-1)k(k+1)(k+2)]. When you add these, every middle piece cancels and only the last one survives.
4. Put in numbers. With n = 10: Σₖ₌₁¹⁰ k(k+1)(k+2) = (10 × 11 × 12 × 13)/4 = 4290. Then 28 Σₖ₌₁¹⁰ 1/aₖ = 28 × 4290/4 = 7 × 4290 = 30030.
5. So. 30030 = 2 × 3 × 5 × 7 × 11 × 13, the product of the first six primes. So m = 6, option (c).

The idea to keep. Given a partial sum, the term is always aₙ = Sₙ - Sₙ₋₁. To add k(k+1)(k+2), attach the next factor (k+3) and divide by 4. The sum telescopes.

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