Cho P = 1/2^2 + 1/3^2 + . . . + 1/2002^2 + 1/2003^2 . Chọn câu đúng

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Cho \[P = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{{2002}^2}}} + \frac{1}{{{{2003}^2}}}\] . Chọn câu đúng

P >1

P >2

P < 1

P < 0

Giải thích

\[P = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{{2002}^2}}} + \frac{1}{{{{2003}^2}}}\]

\[ < \frac{1}{{1.2}} + \frac{1}{{2.3}} + ... + \frac{1}{{2001.2002}} + \frac{1}{{2002.2003}}\]

\[ = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{{2001}} - \frac{1}{{2002}} + \frac{1}{{2002}} - \frac{1}{{2003}}\]

\[ = 1 - \frac{1}{{2003}} = \frac{{2002}}{{2003}} < 1\]

Vậy P < 1

Đáp án cần chọn là: C