(0,5 điểm) Cho C = 3 1 2 .2 2 + 5 2 2 .3 2 + 7 3 2 .4 2 + . . . . + 4047 2023 2 .2024 2 . Chứng minh C < 1.

Hướng dẫn giải

Ta có: \(C = \frac{3}{{{1^2}{{.2}^2}}} + \frac{5}{{{2^2}{{.3}^2}}} + \frac{7}{{{3^2}{{.4}^2}}} + .... + \frac{{4047}}{{{{2023}^2}{{.2024}^2}}}\)

\(C = \frac{{{2^2} - {1^2}}}{{{1^2}{{.2}^2}}} + \frac{{{3^2} - {2^2}}}{{{2^2}{{.3}^2}}} + \frac{{{4^2} - {3^2}}}{{{3^2}{{.4}^2}}} + .... + \frac{{{{2024}^2} - {{2023}^2}}}{{{{2023}^2}{{.2024}^2}}}\)

\(C = \frac{1}{{{1^2}}} - \frac{1}{{{2^2}}} + \frac{1}{{{2^2}}} - \frac{1}{{{3^2}}} + \frac{1}{{{3^2}}} - \frac{1}{{{4^2}}} + .... + \frac{1}{{{{2023}^2}}} - \frac{1}{{{{2024}^2}}}\)

\(C = \frac{1}{{{1^2}}} - \frac{1}{{{{2024}^2}}}\)

\(C = 1 - \frac{1}{{{{2024}^2}}}\)

Nhận thấy \(\frac{1}{{{1^2}}} - \frac{1}{{{{2024}^2}}} < 1\) nên \(C < 1\) (đpcm).