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Ta có :\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}=\)\(\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)\)\(+...+\left(1-\frac{1}{100}\right)\)
=\(\left(1+1+1+....+1\right)\)\(-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(100-1-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
=\(100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)= vế trên (đpcm)
\(S=100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\RightarrowĐPCM\)
Bài 4. Chứng minh rằng
\(\frac{1}{5}< \frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{3}\)
\(\frac{1}{5}< \frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{3}\)
+) Chứng minh: \(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{3}\)
Có: \(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
\(< \frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{3}-\frac{1}{100}< \frac{1}{3}\)
+) Chứng minh \(\frac{1}{5}< \frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
\(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
\(>\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{4}-\frac{1}{101}=\frac{1}{5}+\frac{1}{20}-\frac{1}{101}>\frac{1}{5}\)
\(\frac{1}{5}< \frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{3}\)
Trước hết ta phải chứng minh \(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{3}\)
Ta có \(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)\(< \frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{3}-\frac{1}{100}< \frac{1}{3}\)
Sau đó chứng minh \(\frac{1}{5}< \frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
\(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)\(>\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{4}-\frac{1}{101}=\frac{1}{5}+\frac{1}{20}-\frac{1}{101}>\frac{1}{5}\)
Vậy .................
Ta có: 1/2^2 < 1/1.2
1/3^2 < 1/2.3
.........................
.......................................
1/100^2 < 1/99.100
Ta có: 1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 1/1.2 + 1/2.3 + 1/3.4 + ...... + 1/99.100
1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 1 - 1/100
1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 99/100 < 3/4
1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 3/4
Ta có: 1/2^2 < 1/1.2
1/3^2 < 1/2.3
.........................
.......................................
1/100^2 < 1/99.100
Ta có: 1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 1/1.2 + 1/2.3 + 1/3.4 + ...... + 1/99.100
1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 1 - 1/100
1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 99/100 < 3/4
1/2^2 + 1/3^2 + 1/4^2 +......+1/100^2 < 3/4
Ta có:
\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Mà \(\frac{49}{100}< \frac{1}{2}\)
Vậy \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Ta có:\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)(1)
Xét\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)(2)
Mà\(\frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)(3)
Từ (1), (2), (3)\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\left(đpcm\right)\)
Vậy...
Linz
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2010.2011}\)\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}=1-\frac{1}{2011}=\frac{2010}{2011}>\frac{2010}{2680}=\frac{3}{4}\)
Hình như có gì đó sai sai :')
Đc lém Min đúng lúc tui đang định đăng câu ó
\(Ta\) \(có\) \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{256}\)
\(Vì\) \(1>\frac{1}{256},\frac{1}{2}>\frac{1}{256},....,\frac{1}{255}>\frac{1}{256},\frac{1}{256}=\frac{1}{256}\)
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{256}>\frac{1}{256}+\frac{1}{256}+...+\frac{1}{256}\)
\(=\frac{1}{256}.256=1\)\(< 5\)