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3 tháng 10 2017

\(linh=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+....+\dfrac{100}{5^{100}}\)

\(5linh=5\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{100}{5^{100}}\right)\)

\(5linh=1+\dfrac{2}{5}+\dfrac{3}{5^2}+\dfrac{4}{5^3}+...+\dfrac{100}{5^{99}}\)

\(5linh-linh=\left(1+\dfrac{2}{5}+\dfrac{3}{5^2}+\dfrac{4}{5^3}+...+\dfrac{100}{5^{99}}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{100}{5^{100}}\right)\)

\(4linh=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{99}}-\dfrac{100}{5^{100}}\)

Đặt:

\(linh_2=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{99}}\)

\(5linh_2=5\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{99}}\right)\)

\(5linh_2=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{98}}\)

\(5linh_2-linh_2=\left(5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{98}}\right)-\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{99}}\right)\)

\(4linh_2=5-\dfrac{1}{5^{99}}\)

\(linh=\dfrac{5}{4}-\dfrac{1}{5^{99}.4}\)

Nên \(4linh=\dfrac{5}{4}-\dfrac{1}{5^{99}.4}-\dfrac{100}{5^{100}}\)

Khi đó \(linh=\dfrac{5}{16}-\dfrac{1}{5^{99}.16}-\dfrac{100}{5^{100}.4}\)

3 tháng 10 2017

Bài này bn dùng tính tổng xích ma trên máy tính:

\(\sum\limits^{100}_{x=1}\left(\dfrac{X}{5^X}\right)\)

Kết quả: 5/16

11 tháng 2 2018

\(5D=1+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-\dfrac{1}{5^5}+...+\dfrac{1}{6.5^{99}}\)

\(6D=\dfrac{5^{100}-1}{5^{100}}+\dfrac{1}{6.5^{100}}\)

\(D=\dfrac{\dfrac{5^{100}-1}{5^{100}}+\dfrac{1}{36.5^{100}}}{6}\)

14 tháng 1 2021

\(2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+...+\dfrac{100}{2^{99}}\)

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

Đặt \(B=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}\)

=> \(2B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{98}}\)

=> \(B=\dfrac{1}{2^2}-\dfrac{1}{2^{99}}\)

=> \(A=1+\dfrac{3}{2^2}+\dfrac{1}{2^2}-\dfrac{100}{2^{100}}-\dfrac{1}{2^{99}}\)

=> \(A=2-\dfrac{102}{2^{100}}< 2\)

5 tháng 5 2022

\(=\left(40^1_4-25^1_4\right):\left(\dfrac{5}{7}-\dfrac{5}{7}\right)-\dfrac{1}{2021}\)

\(=15:0-\dfrac{1}{2021}\)

\(=0-\dfrac{1}{2021}\)

\(=\dfrac{-1}{2021}\)

 

 

21 tháng 4 2018

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

\(\dfrac{1}{2}\cdot A=\dfrac{1}{2}+\dfrac{3}{2^4}+...+\dfrac{100}{2^{101}}\)

\(A-\dfrac{A}{2}=\dfrac{1}{2A}=\dfrac{1}{2}+\dfrac{3}{2^3}+...+\dfrac{100}{2^{101}}\)

\(\left[\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right]-\dfrac{100}{2^{101}}\) (do 3/2^3=1/2^2+1/2^3)

\(\left[1-\left(\dfrac{1}{2}\right)^{101}\right]\left(1-\dfrac{1}{2}\right)-\dfrac{100}{2^{101}}\)

\(\left(\dfrac{2^{101-1}}{2^{100}}\right)-\dfrac{100}{2^{101}}\)

\(\Rightarrow A=\dfrac{\dfrac{\left(2^{101-1}\right)}{2^{99}-100}}{2^{100}}\)

22 tháng 5 2017

Giải:

\(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\)

\(\dfrac{1}{2}A=\dfrac{1}{2}+\dfrac{3}{2^4}+\dfrac{4}{2^5}+...+\dfrac{99}{2^{100}}+\dfrac{100}{2^{101}}\)

\(A-\dfrac{A}{2}=\dfrac{1}{2A}=\dfrac{1}{2}+\dfrac{3}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{100}}-\dfrac{100}{2^{101}}\)

\(=\left[\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right]-\dfrac{100}{2^{101}}\) ( Vì \(\dfrac{3}{2^3}=\dfrac{1}{2^2}+\dfrac{1}{2^3}\) )

\(=\dfrac{\left[1-\left(\dfrac{1}{2}\right)^{101}\right]}{\left(1-\dfrac{1}{2}\right)}-\dfrac{100}{2^{101}}\)

\(=\dfrac{\left(2^{101}-1\right)}{2^{100}}-\dfrac{100}{2^{101}}\)

\(\Rightarrow A=\dfrac{\left(2^{101}-1\right)}{2^{99}}-\dfrac{100}{2^{100}}\)

17 tháng 4 2018

2A =2+\(\frac{3}{2^2}\)+\(\frac{4}{2^3}\)+\(\frac{5}{2^4}\)+.....+\(\frac{100}{2^{99}}\)

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

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

\(\Rightarrow\)A=2A-A=1+\(\frac{3}{4}\)+\(\frac{1}{4}\)-\(\frac{101}{2^{99}}\)+\(\frac{100}{2^{100}}\)=2-\(\frac{51}{2^{99}}\)

20 tháng 3 2017

A=1+B

B=\(\Sigma\left(\dfrac{x}{2^x}\right)\)( cho x chạy từ 3 đến 100) =1

=> A=1+B=1+1=2ok

30 tháng 3 2018

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

\(\Leftrightarrow2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+.........+\dfrac{100}{2^{99}}\)

\(\Leftrightarrow2A-A=\left(2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+......+\dfrac{100}{2^{99}}\right)-\left(1+\dfrac{3}{2^3}+.........+\dfrac{100}{2^{100}}\right)\)

\(\Leftrightarrow A=\dfrac{11}{4}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+......+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)

Đặt :

\(H=\dfrac{1}{2^3}+\dfrac{1}{2^4}+......+\dfrac{1}{2^{99}}\)\(\Leftrightarrow A=\dfrac{11}{4}-H-\dfrac{100}{2^{100}}\)

\(\Leftrightarrow2H=\dfrac{1}{2^2}+\dfrac{1}{2^3}+........+\dfrac{1}{2^{98}}\)

\(\Leftrightarrow2H-H=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2^3}+\dfrac{1}{2^4}+.....+\dfrac{1}{2^{99}}\right)\)

\(\Leftrightarrow H=\dfrac{1}{2^2}-\dfrac{1}{2^{99}}\)

\(\Leftrightarrow A=\dfrac{11}{4}+\dfrac{1}{2^2}-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)