C/M công thức tổng quát:\(n^3>n^3-n\Rightarrow\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{n\left(n^2-1\right)}=\frac{1}{\left(n-1\right)n\left(n+1\right)}\)

\(\Rightarrow\frac{1}{n^3}< \frac{1}{\left(n-1\right)n\left(n+1\right)}\)

Đặt \(A=\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\frac{1}{5^3}+.....+\frac{1}{2017^3}\)

Áp dụng vào bài toán,ta được:\(A< \frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+....+\frac{1}{2016\cdot2017\cdot2018}\)

\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+....+\frac{1}{2016\cdot2017}-\frac{1}{2017\cdot2018}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2017\cdot2018}\right)\)

\(=\frac{1}{4}-\frac{1}{2\cdot2017\cdot2018}\)

\(< \frac{1}{2^2}^{ĐPCM}\)

Ta có: 

\(A=\frac{1}{2}+\frac{1}{2^2}+........+\frac{1}{2^{2017}}\)

\(\Rightarrow2A=1+\frac{1}{2}+.........+\frac{1}{2^{2016}}\)

Khi đó: 

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

\(\Rightarrow A=1-\frac{1}{2^{2017}}\)

\(\Rightarrow A=\frac{2^{2017}-1}{2^{2017}}\)

\(\Rightarrow A< 1\)

VẬy: A < 1

Ta có:                                                                       1/2+1/2^2+...+1/2^2017<1/1.2+1/2.3+...+1/2016.2017

1/2<1/1.2

1/2^2<1/2.3

..........

1/2^2017<1/2016.2017

Câu 8( Mình không viết đè nữa nha)

a)   2-1/1.2 + 3-2/2.3 + 4-3/3.4 +…..+ 100-99/99.100

=  1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 +…..+ 1/99 – 1/100

=  1 – 1/100 < 1

=   99/100 < 1

    Vậy A< 1

cảm ơn bạn nhiều

Phần C đề thiếu

\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)

\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)

\(\Rightarrow4D=3-\frac{203}{3^{100}}\)

\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)

sửa rồi nhá bn

a/

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(A=2A-A=1-\frac{1}{2^{100}}< 1\)

b/

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)

\(2B=3B-B=1-\frac{1}{3^{2019}}\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2019}}< \frac{1}{2}\)