If 1/α+1 + 1/α+2 + 1/α+3 +...+ 1/α+1012 − (1/2×1 + 1/4×3 + 1/6×5 +....+ 1/2024 × 1/2023) = 1/2024,

Question

If \(\frac1{\alpha + 1}+ \frac1{\alpha + 2} + \frac 1{ \alpha + 3} + ... + \frac 1{ \alpha + 1012} - \left( \frac 1 {2 \times 1} + \frac 1{4 \times 3} + \frac1{6 \times 5} + ....+ \frac 1{2024} \times\frac 1{2023} \right) = \frac 1 {2024}\), then α is equal to

(1) 2012

(2) 1012

(3) 1011

(4) 506

Answer 1

Correct option is (3) 1011