(0,5 điểm) Chứng minh rằng M = 1 2 2 + 1 3 2 + 1 4 2 + . . . + 1 50 2 < 1 .

Hướng dẫn giải

Ta có: \(\frac{1}{{{2^2}}} = \frac{1}{{2.2}} < \frac{1}{{1.2}}\)

\(\frac{1}{{{3^2}}} = \frac{1}{{3.3}} < \frac{1}{{2.3}}\)

\(\frac{1}{{{4^2}}} = \frac{1}{{4.4}} < \frac{1}{{3.4}}\)

….

\(\frac{1}{{{{50}^2}}} = \frac{1}{{50.50}} < \frac{1}{{49.50}}\)

Do đó, \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + .... + \frac{1}{{{{50}^2}}} < \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + .... + \frac{1}{{49.50}}\)

Suy ra \(M < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + .... + \frac{1}{{49}} - \frac{1}{{50}}\) hay \(M < 1 - \frac{1}{{50}}\).

Như vậy \(M < \frac{{49}}{{50}}\) hay \(M < 1.\)