ơi

bé

không

Bài 2 :

\(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+............+\dfrac{2017}{4^{2017}}\)

\(\Leftrightarrow4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...........+\dfrac{2017}{4^{2016}}\)

\(\Leftrightarrow4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+..........+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+..........+\dfrac{2017}{4^{2017}}\right)\)

\(\Leftrightarrow3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+.........+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)

Đặt :

\(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2016}}\)

\(\Leftrightarrow4A=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2015}}\)

\(\Leftrightarrow4A-A=\left(4+1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2016}}\right)\)

\(\Leftrightarrow3A=4-\dfrac{1}{4^{2016}}\)

\(\Leftrightarrow D=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}\)

\(\Leftrightarrow3S=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}-\dfrac{2017}{4^{2016}}\)

\(\Leftrightarrow3S< \dfrac{4}{3}\)

\(\Leftrightarrow S< \dfrac{4}{9}\)

\(\Leftrightarrow S< \dfrac{1}{2}\rightarrowđpcm\)

\(A=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\) ( A cho đẹp :v)

\(4A=4\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)

\(4A=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\)

\(4A-A=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)\(3A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)

Đặt:

\(M=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\)

\(4M=4\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)

\(4M=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\)

\(4M-M=\left(4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3M=4-\dfrac{1}{4^{2016}}\)

\(M=\dfrac{4}{3}-\dfrac{1}{4^{2016}}\)

Thay M vào A ta có:

\(A=\dfrac{4}{9}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2017}}\)

\(\Rightarrow A< \dfrac{1}{2}\Rightarrowđpcm\)

Đặt A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

A=\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)

Ta thấy :

\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\)

\(\dfrac{1}{100.100}< \dfrac{1}{99.100}\)

\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

Nhân xét :

\(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};\)

\(...;\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}+...+\)

\(\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow A< 1-\dfrac{1}{100}\)

\(\Rightarrow A< \dfrac{99}{100}\)

Vì \(A< \dfrac{99}{100}< 1\)

\(\Rightarrow A< 1\)

Bài 1)

Đặt \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\)
Ta thấy:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4};....;\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\)\(\Rightarrow\) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\) < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
\(\Rightarrow\) A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow\) A < \(1-\dfrac{1}{100}\) < 1 \(\Rightarrow\) A < 1

Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\)< 1

Bài 1:

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1\left(đpcm\right)\)

Bài 2: bạn xem lại xem có phải \(S=3^0+3+3^2+...+3^{2002}\) không nhé!

1/2^2+1/3^2+...+1/50^2<1/1*2+1/2*3*+...+1/49*50

=1/1-1/2+1/2-1/3+...+1/49-1/50<1

=>S<1+1=2

\(S=\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)

=>\(S< =\dfrac{1}{4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)

=>\(S< =\dfrac{1}{4}\cdot\left(1-\dfrac{1}{n}\right)=\dfrac{1}{4}\cdot\dfrac{n-1}{n}< =\dfrac{1}{4}\)

Ta có:
1/2^2 > 1/2.3
1/3^2 > 1/3.4
...
1/10^2 > 1/10.11
-> Cộng dọc theo vế ta có:
1/2^2+1/3^2+...+1/10^2 > 1/2.3+1/3.4+...+1/10.11
                                         = 1/2-1/3+1/3-1/4+...+1/10-1/11 

                                         = 1/2 - 1/11 = 9/22  (đpcm)         

giúp mk với ;-;"

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

Ta có: \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\)

    \(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

    \(=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)

\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)

\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)

......

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Rightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)