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

3A = 1 + \(\frac{2}{3}\) + \(\frac{3}{3^2}\) + \(\frac{4}{3^3}\)  +...+ \(\frac{100}{3^{99}}\) 

\(\Rightarrow\) 3A - A = 1+ \(\left(\frac{2}{3}-\frac{1}{3}\right)\) + \(\left(\frac{3}{3^2}-\frac{2}{3^2}\right)\) + ... + \(\left(\frac{100}{3^{99}}-\frac{99}{3^{99}}\right)\)   - \(\frac{100}{3^{100}}\)  

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

Đặt B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{99}}\)

\(\Rightarrow\) 3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\) 

\(\Rightarrow\) 2B = \(1-\frac{1}{3^{99}}\)  

\(\Rightarrow\) \(B=\left(1-\frac{1}{3^{99}}\right):2\)   

Thay 2A = 1 + \(\frac{1}{2}\) - \(\left(1-\frac{2}{3^{99}}\right)\)  - \(\frac{100}{3^{100}}\)   < 1 + \(\frac{1}{2}\) = \(\frac{3}{2}\)  

Vậy A < \(\frac{3}{4}\) 

Vậy:...........

Đặt :

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

\(\Leftrightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+.....+\frac{100}{3^{99}}\)

\(\Leftrightarrow3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+....+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+....+\frac{100}{3^{100}}\right)\)

\(\Leftrightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt : \(H=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\) \(\Leftrightarrow2A=H-\frac{100}{3^{100}}\)

\(\Leftrightarrow3H=3+1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}\)

\(\Leftrightarrow3H-H=\left(4+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\right)\)

\(\Leftrightarrow2H=3-\frac{1}{3^{99}}\)

\(\Leftrightarrow H=\frac{3-\frac{1}{99}}{2}\)

\(\Leftrightarrow2A=\frac{3-\frac{1}{3^{99}}}{2}-\frac{100}{3^{100}}\)

\(\Leftrightarrow A=\frac{1-\frac{1}{3^{99}}}{2}-\frac{100}{2.3^{100}}\)

\(\Leftrightarrow A< \frac{3}{4}\left(đpcm\right)\)

Test câu trả lời

Chúc bạn học tốt!

hình như đề sai bởi vì trong dãy số có số 4/4^3

\(\frac{4}{3^4}\)moi dung

đề sai a

a/ Ta có

\(200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)\)

\(=1+2\left(1-\frac{1}{3}\right)+2\left(1-\frac{1}{4}\right)+...+2\left(1-\frac{1}{100}\right)\)

\(=1+2\left(\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)\)

\(=2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)

Thế lại bài toán ta được:

\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)

\(=\frac{2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}=2\)

b/ Ta có: 

A - B\(=\frac{-21}{10^{2016}}+\frac{12}{10^{2016}}+\frac{21}{10^{2017}}-\frac{12}{10^{2017}}\)

\(=\frac{9}{10^{2017}}-\frac{9}{10^{2016}}< 0\)

Vậy A < B

cảm ơn bạn

Bài 1:
Ta có:

\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)

\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Mà \(\frac{99}{100}< 1\)

\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)

Có phải ở sách NCPT ko bn

Bài 2: Đặt \(B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)

\(3B=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)

\(3B-B=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)

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

\(6B=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(6B-2B=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(4B=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)

\(4B=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)

\(4B=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)

\(4B=3-\frac{203}{3^{100}}< 3\)

\(B< \frac{3}{4}\left(đpcm\right)\)

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