Let for n = 1, 2, ......, 50, Sₙ be the sum of the infinite geometric progression whose first term is n² and whose common ratio is 1/(n + 1)². Then the value of
1/26 + Σₙ₌₁⁵⁰ ( Sₙ + 2/(n + 1) - n - 1 ) is equal to ___________.

Solution

Maths sir · 5 steps

Split Sₙ into a polynomial plus a fraction, and the fractions telescope away beautifully.

1. Find S_n. The common ratio 1/(n+1)² is less than 1, so the infinite sum is a/(1-r). Since (n+1)² - 1 = n(n+2), this gives Sₙ = n²(n+1)²/n(n+2) = n(n+1)²/(n+2).
2. Divide it out. Dividing n³ + 2n² + n by n+2 gives quotient n² and remainder n. Writing n/(n+2) = 1 - 2/(n+2) gives Sₙ = n² + 1 - 2/(n+2).

   Sₙ + 2/(n+1) - n - 1 = (n² - n) + 2(1/(n+1) - 1/(n+2))

3. Polynomial part. Σₙ₌₁⁵⁰ n² = (50 × 51 × 101)/6 = 42925 and Σₙ₌₁⁵⁰ n = 1275. Their difference is 41650.
4. Telescoping part. In the bracket, each 1/(n+2) cancels the next term's 1/(n+1). Only the first and last pieces survive: 2(1/2 - 1/52) = 1 - 1/26.
5. So. The extra 1/26 cancels the -1/26. The value is 1/26 + 41650 + 1 - 1/26 = 41651.

The idea to keep. When a sum mixes a polynomial with fractions like 1/(n+1) and 1/(n+2), divide out the polynomial first. The leftover fractions usually telescope.

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