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p=1-1/2.5-1/3.5-1/1.3-1/4.7-1/2.3-1/3.7
p=1-(1/2.1/5-1/3.1/5)-(1/1.1/3-1/2.1/3)-(1/4.1/7-1/3.1/7)
p=1-(1/5.(1/2-1/3))-(1/3.(1-1/2))-(1/7.(1/4-1/3)
p=1-(1/5.1/6)-(1/3.1/2)-(1/7.-1/12)
p=1-1/30-1/6+1/84
p=341/420
\(\left(\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}\right)-\left(\frac{-5}{4}+\frac{6}{11}-\frac{48}{49}\right)=\left(\frac{-1}{4}-\frac{16}{11}\right)-\left(-\frac{31}{44}-\frac{48}{49}\right)=-\frac{1}{4}-\frac{16}{11}+\frac{31}{44}+\frac{48}{49}=-\frac{1}{49}\)
Ta có:
\(\frac{A}{2}=\frac{3^3}{2}-\frac{5^3}{6}+\frac{7^3}{12}-\frac{9^3}{20}+\frac{11^3}{30}-\frac{13^3}{42}+\frac{15^3}{56}-\frac{17^3}{72}+...+\frac{199^3}{9900}\)
\(=3^2.\left(1+\frac{1}{2}\right)-5^2.\left(\frac{1}{2}+\frac{1}{3}\right)+7^2.\left(\frac{1}{3}+\frac{1}{4}\right)-9^2.\left(\frac{1}{4}+\frac{1}{5}\right)+...+199^2.\left(\frac{1}{99}+\frac{1}{100}\right)\)
\(=3^2+\left(\frac{3^2}{2}-\frac{5^2}{2}\right)-\left(\frac{5^2}{3}-\frac{7^2}{3}\right)+\left(\frac{7^2}{4}-\frac{9^2}{4}\right)-\left(\frac{9^2}{5}-\frac{11^2}{5}\right)+...+\left(\frac{197^2}{99}-\frac{199^2}{99}\right)+\frac{199^2}{100}\)
\(=3^2-8+8-8+...+8+\frac{199^2}{100}=3^2+\frac{199^2}{100}< 3^2+\frac{199.200}{100}=9+398=407\)
\(\Rightarrow A< 407.2=814\)
ta gọi biểu thức trên là B có
2B=2.(\(\frac{1}{6}\)+\(\frac{1}{10}\)+\(\frac{1}{15}\)+....+\(\frac{1}{4950}\))
2B=\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+......+\frac{1}{9900}\)
2B=\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.......+\frac{1}{99.100}\)
2B=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)+.....+\(\frac{1}{99}-\frac{1}{100}\)
2B=\(\frac{1}{3}-\frac{1}{100}\)
2B=\(\frac{100-3}{300}\)
B=\(\frac{97}{300}\): 2
B=\(\frac{97}{300}.\frac{1}{2}\)
B=\(\frac{97}{600}\)
