S)\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
S)=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
S)=\(\frac{1}{3}-\frac{1}{101}\)
S)=\(\frac{98}{303}\)
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ta có:
2/3.5 = 1/3 ‐ 1/5
tương tự:
2/5.7 = 1/5 ‐ 1/7
2/7.9 = 1/7 ‐ 1/9
...........
2/99.101 = 1/99 + 1/101
=> B = 1/3 ‐ 1/5 + 1/5 ‐ 1/7 + 1/7 ‐ 1/9 +...+1/99 ‐ 1/101
= 1/3 ‐ 1/101
= 98/303
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