Question

Chứng minh rằng:

\(\frac{3}{1^2\times2^2}+\frac{5}{2^2\times3^2}+\frac{7}{3^2\times4^2}+......+\frac{19}{9^2\times10^2}\) \(<1\) 

 

Answer 1

Dưới tử mik tính ra thôi. VD: 1² . 2² = 1.4; 2².3² = 4.9 các tử sau tương tự

\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)

= \(\frac{4-1}{1.4}+\frac{9-4}{4.9}+\frac{16-9}{9.16}+...+\frac{100-81}{81.100}\)

= \(\frac{4}{1.4}-\frac{1}{1.4}+\frac{9}{4.9}-\frac{4}{4.9}+\frac{16}{9.16}-\frac{9}{9.16}\)+.....+\(\frac{100}{81.100}-\frac{81}{81.100}\)

= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)

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