(0,5 điểm) Cho 1 2 A = 1 3 2 + 1 5 2 + 1 7 2 + 1 9 2 + . . . + 1 2025 2 . Chứng minh rằng A < 506 1013 .

Hướng dẫn giải

Ta có: \(\frac{1}{2}A = \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \frac{1}{{{7^2}}} + \frac{1}{{{9^2}}} + ... + \frac{1}{{{{2025}^2}}}\)

Suy ra \(A = \frac{2}{{{3^2}}} + \frac{2}{{{5^2}}} + \frac{2}{{{7^2}}} + \frac{2}{{{9^2}}} + ... + \frac{2}{{{{2025}^2}}}\)

Nhận thấy \(\frac{2}{{{3^2}}} = \frac{2}{9} < \frac{2}{8} = \frac{2}{{2.4}}\)

\(\frac{2}{{{5^2}}} = \frac{2}{{25}} < \frac{2}{{24}} = \frac{2}{{4.6}}\)

…….

\(\frac{2}{{{{2025}^2}}} = \frac{2}{{2025.2025}} < \frac{2}{{2024.2026}}\).

Cộng theo vế, ta được:

\(\frac{2}{{{3^2}}} + \frac{2}{{{5^2}}} + \frac{2}{{{7^2}}} + \frac{2}{{{9^2}}} + ... + \frac{2}{{{{2025}^2}}} < \frac{2}{{2.4}} + \frac{2}{{4.6}} + ... + \frac{2}{{2024.2026}}\)

\(A < \frac{1}{2} - \frac{1}{4} + \frac{1}{4} - \frac{1}{6} + ... + \frac{1}{{2024}} - \frac{1}{{2026}}\)

\(A < \frac{1}{2} - \frac{1}{{2026}}\)

\(A < \frac{{506}}{{1013}}\).

Vậy \(A < \frac{{506}}{{1013}}\).