Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{97.99}+\frac{1}{98.100}\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+...+\frac{1}{2}.\left(\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}.\left(\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{97}+\frac{1}{98}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-...-\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{99}-\frac{1}{100}\right)< \frac{1}{2}.\left(1+\frac{1}{2}\right)=\frac{3}{4}\)
\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{ }\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\times\frac{58}{45}=\frac{29}{45}\)
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}=\dfrac{49}{99}>\dfrac{49}{100}=A\)
Lời giải:
Xét thừa số tổng quát $1+\frac{1}{n(n+2)}=\frac{n(n+2)+1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$
Khi đó:
$1+\frac{1}{1.3}=\frac{2^2}{1.3}$
$1+\frac{1}{2.4}=\frac{3^2}{2.4}$
.........
$1+\frac{1}{99.101}=\frac{100^2}{99.101}$
Khi đó:
$A=\frac{2^2.3^2.4^2......100^2}{(1.3).(2.4).(3.5)....(99.101)}$
$=\frac{(2.3.4...100)(2.3.4...100)}{(1.2.3...99)(3.4.5...101)}$
$=\frac{2.3.4...100}{1.2.3..99}.\frac{2.3.4...100}{3.4.5..101}$
$=100.\frac{2}{101}=\frac{200}{101}$
\(A=\frac{1}{1.3}+\frac{1}{3.5}+..+\frac{1}{97.99}\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(A=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
\(\Leftrightarrow A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
\(\Leftrightarrow A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
\(S=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{97.99}+\frac{1}{98.100}\)
\(S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}\)
\(S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)
\(S< 1-\frac{1}{99}< 1\)
=> S < 1
Cảm ơn bạn nhé