cho mk hỏi:
\(E=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
Y
28 tháng 3 2019
\(3B=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(B=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4B=3B+B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
+ Đặt \(M=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(3M=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(\Rightarrow4M=3M+M=3-\frac{1}{3^{99}}\)
\(\Rightarrow M=\frac{3}{4}-\frac{1}{3^{99}\cdot4}\)
\(\Rightarrow4B=M-\frac{100}{3^{100}}=\frac{3}{4}-\frac{1}{3^{99}\cdot4}-\frac{100}{3^{100}}\)
\(\Rightarrow B=\frac{3}{16}-\frac{1}{3^{99}\cdot16}-\frac{100}{3^{100}\cdot4}\) \(\Rightarrow B< \frac{3}{16}\)
Y
28 tháng 3 2019
a) \(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
\(\Rightarrow3A=2A+A=1-\frac{1}{2^6}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{2^6\cdot3}< \frac{1}{3}\) ( đpcm )
31 tháng 8 2022
a: =>x-8/5=1/20-1/10=-1/20
=>x=-0,05+1,6=1,55
b: =>x-3/2=4/3 hoặc x-3/2=-4/3
=>x=17/6 hoặc x=1/6
c: =>\(\left|x-\dfrac{1}{3}\right|=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{2}{3}=\dfrac{35}{12}\)
=>x-1/3=35/12 hoặc x-1/3=-35/12
=>x=39/12=13/4 hoặc x=-31/12
d: =>|x-5/8|=3/4
=>x-5/8=3/4 hoặc x-5/8=-3/4
=>x=11/8 hoặc x=-1/8
KN
25 tháng 3 2019
a) \(\frac{x-3}{3}-1=\frac{x}{-4}\)
\(\Leftrightarrow\frac{x-3}{3}-\frac{3}{3}=\frac{x}{-4}\)
\(\Leftrightarrow\frac{x-6}{3}=\frac{x}{-4}\)
\(\Leftrightarrow-4\left(x-6\right)=3x\)
\(\Leftrightarrow-4x+24=3x\)
\(\Leftrightarrow24=3x+4x\)
\(\Leftrightarrow7x=24\)
\(\Leftrightarrow x=\frac{24}{7}\)
KN
25 tháng 3 2019
b) \(\frac{5}{8}-\left(x-\frac{1}{2}\right)=\frac{-3}{4}\)
\(\Leftrightarrow\frac{5}{8}-x+\frac{1}{2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{5}{8}+\frac{4}{8}-x=\frac{-3}{4}\)
\(\Leftrightarrow\frac{9}{8}-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{9}{8}+\frac{3}{4}\)
\(\Leftrightarrow x=\frac{15}{8}\)
AK
9 tháng 5 2018
Đặt \(A=\frac{1}{9}+\frac{2}{8}+...+\frac{8}{2}+\frac{9}{1}\)
\(\Rightarrow A=\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+...+\frac{8}{2}+\left(1+1+...+1\right)\left(9cs1\right)\)
\(\Rightarrow A=\left(\frac{1}{9}+1\right)+\left(\frac{2}{8}+1\right)+...+\left(\frac{8}{2}+1\right)+1\)
\(\Rightarrow A=\frac{10}{9}+\frac{10}{8}+...+\frac{10}{2}+\frac{10}{10}\)
\(\Rightarrow A=10.\left(\frac{1}{2}+...+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)
Mà \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{10}\right).x=A\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right).x=\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right).10\)
\(\Rightarrow x=10\)
Vậy \(x=10\)
D
8 tháng 9 2019
a) MC :24
\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}=\frac{1\times8+3\times3-7\times2}{24}=\frac{3}{24}=\frac{1}{8}\)
b)MC : 56
\(\frac{3}{14}+\frac{5}{8}-\frac{1}{2}=\frac{3\times4+5\times7-1\times28}{56}=\frac{19}{56}\)
c) MC: 36
\(\frac{1}{4}-\frac{2}{3}-\frac{11}{18}=\frac{1\times9-2\times12-11\times2}{36}=\frac{-37}{36}\)
d) MC: 312
\(\frac{1}{4}+\frac{5}{12}-\frac{1}{13}-\frac{7}{8}=\frac{1\times78+5\times26-1\times24-7\times39}{312}=\frac{-89}{312}\)
\(E=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\\ 3E=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\\ 3E-E=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\\ 2E=1-\frac{1}{3^8}\\ E=\frac{1-\frac{1}{3^8}}{2}\)
thank you!