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Ta có:
A=\(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)
A<\(1+\dfrac{1}{2.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
A<\(1+\dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
A<\(\dfrac{5}{4}\)+\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{99}-\dfrac{1}{100}\)
A<\(\dfrac{5}{4}+\dfrac{1}{2}-\dfrac{1}{100}\)
A<\(\dfrac{5}{4}+\dfrac{49}{100}\)
A<\(\dfrac{174}{100}\)<\(\dfrac{7}{4}\)
=>A<\(\dfrac{7}{4}\)
Tick giùm mink nha :D
Giải:
Ta có:
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\)
Đặt \(A=\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(B=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
\(A=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\dfrac{1}{4}-\dfrac{1}{100}\)
\(A=\dfrac{6}{25}\)
Mà \(\dfrac{1}{6}< \dfrac{6}{25}< \dfrac{1}{4}\)
Ta lại có \(A< \dfrac{6}{25}\)
Vậy \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
1/5^2< 1/4.5=1/4-1/5
1/6^2<1/5.6=1/5-1/6
..
1/99^2<1/98.99=1/98-1/99
1/100^2<1/99.100=1/99-1/100
Cộng vế theo vế, đơn giản:
=> 1/5^2+1/6^2+...+1/100^2< 1/4 -1/100<1/4
**
1/5^2> 1/5.6=1/5-1/6
1/6^2>1/6.7=1/6-1/7
..
1/99^2>1/99.100=1/99-1/100
1/100^2>1/100.101=1/100-1/101
Cộng vế theo vế, đơn giản:
=> 1/5^2+1/6^2+...+1/100^2>1/5 -1/101=96/505>1/6
Vậy:
1/6<1/5^2+1/6^2+...+1/100^2<1/4
Ta có:\(A=\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{98}{99}\)
\(A< \dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\dfrac{7}{8}\cdot...\cdot\dfrac{99}{100}\)
\(\Rightarrow A^2< \dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}\cdot\dfrac{7}{8}\cdot...\cdot\dfrac{98}{99}\cdot\dfrac{99}{100}\)
\(A^2< \dfrac{2}{100}=\dfrac{1}{50}\)
Mà \(\dfrac{1}{50}< \dfrac{1}{49}\)
\(\Rightarrow A^2< \dfrac{1}{49}\)
\(\Rightarrow A< \dfrac{1}{7}\left(đpcm\right)\)
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B< 1-\dfrac{1}{8}=\dfrac{7}{8}< 1\)
mink nhanh nhất đó bạn,
ta có :
\(\dfrac{1}{2^2}< \dfrac{1}{1\times2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\times3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3\times4}\)
. . . . . . .
\(\dfrac{1}{8^2}< \dfrac{1}{7\times8}\)
_________________________________
\(\Rightarrow\)\(B< \)\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{7.8}\right)\)
\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{7}-\dfrac{1}{8}\)
\(\Rightarrow B< 1-\dfrac{1}{8}\)
\(\Rightarrow B< 1\)
\(\Rightarrowđpcm\)
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
......
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
Ta có: \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
1/4^2 + 1/5^2 +... + 1/100^2 < 1/3.4 + 1/4.5 +...+ 1/99.100
A=1/3 - 1/4 + 1/4 - 1/5 +...+ 1/99 - 1/100
=1/3 - 1/100 < 1/3
2,
\(M=\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\) =\(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)
\(=\dfrac{3}{4}\)
a, Ta có: \(\dfrac{1}{n}.\dfrac{1}{n+4}=\dfrac{1}{n.\left(n+4\right)}=\dfrac{1}{4}.\dfrac{4}{n.\left(n+1\right)}=\dfrac{1}{4}.\left(\dfrac{1}{n}-\dfrac{1}{n+4}\right)\)
Vậy \(\dfrac{1}{n}.\dfrac{1}{n+1}=\dfrac{1}{4}.\left(\dfrac{1}{n}-\dfrac{1}{n+4}\right)\)
b, \(A=\dfrac{4}{3}.\dfrac{4}{7}+\dfrac{4}{7}.\dfrac{4}{11}+...+\dfrac{4}{95}.\dfrac{4}{99}=4.\left(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{95.99}\right)\)
\(=4.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{99}\right)\)
\(=4.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)=4.\dfrac{32}{99}=\dfrac{128}{99}\)
Vậy \(A=\dfrac{128}{99}\)
Lời giải:
Đặt \(A=\frac{1}{7^2}-\frac{1}{7^4}+....+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
\(7^2A=1-\frac{1}{7^2}+....+\frac{1}{7^{4n-4}}-\frac{1}{7^{4n-2}}+...+\frac{1}{7^{96}}-\frac{1}{7^{98}}\)
\(\Rightarrow A+7^2A=1-\frac{1}{7^{100}}\Rightarrow 50A=1-\frac{1}{7^{100}}<1\)
$\Rightarrow A< \frac{1}{50}$