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

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)

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

cunasai

Tham khảo nha bạn :

Câu hỏi của Trần Minh Hưng - Toán lớp | Học trực tuyến

Đặt A= 200- (3+\(\frac{2}{3}+\frac{2}{4}+.....+\frac{2}{100}\))

         =\(197-\frac{2}{3}-\frac{2}{4}-....-\frac{2}{100}\)

        =\(\frac{197.2}{2}-\frac{2}{3}-\frac{2}{4}-....-\frac{2}{100}\)

        =\(2.\left(\frac{196+1}{2}-\frac{1}{3}-\frac{1}{4}-.....-\frac{1}{100}\right)\)

        =\(2\left(\frac{196}{2}+\frac{1}{2}-\frac{1}{3}-.....-\frac{1}{100}\right)\)

         =\(2\left(98+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-.....-\frac{1}{100}\right)\)

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

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

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

M = 1/3 + 2/3² + 3/3³ + 4/3^4 + ... + 100/3^100 

=> 3M= 1 + 2/3 + 3/3² + 4/3³ + .... + 100/3^99

=> 3M-M = 1 + ﴾2/3 ‐ 1/3﴿ + ﴾3/3² ‐ 2/3²﴿ +...+ ﴾100/3^99 ‐ 99/3^99﴿ ‐ 100/3^100 

=> 2M= 1+ 1/3 + 1/3² + 1/3³ +...+ 1/3^99 ‐ 100/3^100 

Đặt N = 1/3 + 1/3² + 1/3³ +...+ 1/3^99

=> 3N = 1 + 1/3 + 1/3² + 1/3³ +...+ 1/3^98

=> 2N = 1 ‐ 1/3^99 

=> N = ﴾1 ‐ 1/3^99﴿/2 

Thay vào 2M 

=> 2M= 1+ 1/2 ‐ 1/﴾2x3^99﴿ ‐ 100/3^100 < 1+ 1/2 = 3/2 

=> M < 3/4 

vậy...

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