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\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
+) \(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)>\left(\frac{1}{75}+...+\frac{1}{75}\right)+\left(\frac{1}{100}+...+\frac{1}{100}\right)\)
=> \(A>\frac{25}{75}+\frac{25}{100}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
+) \(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)<\left(\frac{1}{50}+...+\frac{1}{50}\right)+\left(\frac{1}{75}+...+\frac{1}{75}\right)\)
=> \(A<\frac{25}{50}+\frac{25}{100}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}<\frac{5}{6}\)
Vậy...
Chứng minh : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\): như câu trên
+) \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{9900}\)
\(A=\left(\frac{1}{2}+\frac{1}{12}\right)+\left(\frac{1}{30}+...+\frac{1}{9900}\right)>\frac{1}{2}+\frac{1}{12}.\)
\(\Rightarrow A>\frac{1}{2}+\frac{1}{12}\)
\(\Rightarrow A>\frac{7}{12}\left(1\right).\)
+) \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\left(1-\frac{1}{2}+\frac{1}{3}\right)-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{5}{6}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}< \frac{5}{6}\)
\(\Rightarrow A< \frac{5}{6}\left(2\right).\)
Từ \(\left(1\right)và\left(2\right)\Rightarrow\frac{7}{12}< A< \frac{5}{6}\left(đpcm\right).\)
Chúc bạn học tốt!
Ta thấy : A = 1 - 1/2 + 1/3 - 1/4 +...+ 1/99 +1/100
A = 1 + 1/2 + 1/3 + 1/4 +...+ 1/99 + 1/100 - 2. (1/2 + 1/4 +1/6 +...+ 1/100)
A = 1 + 1/2 + 1/3 +1/4 +...+ 1/99 + 1/100 - (1 + 1/2 + 1/3 +...+1/50)
A = 1/51 + 1/52 + 1/53 +...+ 1/100
Do đó : A = (1/51 + 1/52 + 1/53 +...+ 1/75) + (1/76 + 1/77 +...+ 1/100)
Ta có : 1/51 > 1/52 > ... > 1/75 ; 1/76 > 1/77 > ... > 1/100 nên :
A > 1/75.25 + 1/100.25 = 1/3 + 1/4 = 7/12
A < 1/51.25 < 1/50.25 + 1/75.25 = 1/2 +1/3 = 5/6
Vậy 7/12 < A <5/6 . ^_^
Câu hỏi của Vũ Thị Kim Oanh - Toán lớp 7 - Học toán với OnlineMath
Tham khảo
A= 1-\(\frac{1}{2}\) +\(\frac{1}{3}\) - \(\frac{1}{4}\) +\(\frac{1}{5}\)- \(\frac{1}{6}\) + ...+ \(\frac{1}{99}\) - \(\frac{1}{100}\)
= 1+ \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\) + \(\frac{1}{6}\) + ...+ \(\frac{1}{99}\) + \(\frac{1}{100}\) - 2 ( \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{6}\) + ...+ \(\frac{1}{100}\) )
= 1+ \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) + ...+ \(\frac{1}{99}\) + \(\frac{1}{100}\)
= \(\frac{1}{51}\) + \(\frac{1}{52}\) +...+ \(\frac{1}{100}\)
= (\(\frac{1}{51}\) + \(\frac{1}{52}\) + ... + \(\frac{1}{75}\) ) + ( \(\frac{1}{76}\) + \(\frac{1}{77}\) + ... + \(\frac{1}{100}\) )
Ta có : \(\frac{1}{51}\) > \(\frac{1}{52}\) > \(\frac{1}{53}\) > ... > \(\frac{1}{75}\)
\(\frac{1}{76}\) > \(\frac{1}{77}\) > \(\frac{1}{78}\) > ... > \(\frac{1}{100}\)
=> A > \(\frac{1}{75}.25\) + \(\frac{1}{100}.25\) = \(\frac{1}{3}\) + \(\frac{1}{4}\) = \(\frac{7}{12}\)
=> A< \(\frac{1}{51}.25\) + \(\frac{1}{75}.25\) < \(\frac{1}{50}.25\) + \(\frac{1}{75}.25\) = \(\frac{1}{2}\) + \(\frac{1}{3}\) = \(\frac{5}{6}\)
Vậy \(\frac{7}{12}\) < A < \(\frac{5}{6}\)
Tick nha