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

Đặt \(C=\dfrac{1}{4}-\dfrac{1}{4^2}+...-\dfrac{1}{4^{100}}\)

\(\Leftrightarrow C\cdot\dfrac{1}{4}=\dfrac{1}{4^2}-\dfrac{1}{4^3}+...-\dfrac{1}{4^{101}}\)

\(\Leftrightarrow C\cdot\dfrac{-3}{4}=\dfrac{-1}{4^{101}}-\dfrac{1}{4}=\dfrac{-1-4^{100}}{4^{101}}\)

\(\Leftrightarrow C=\dfrac{-4^{100}-1}{4^{101}}\cdot\dfrac{-4}{3}=\dfrac{4^{100}+1}{3\cdot4^{100}}\)

\(\Leftrightarrow B=\dfrac{-4^{100}-1}{4^{100}}\)

16 tháng 4 2017

a)Ta có :

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

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

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

\(3A=1-\dfrac{1}{4^{100}}\)

\(\Rightarrow A=\dfrac{1-\dfrac{1}{4^{100}}}{3}\)

~ Chúc bn học tốt ~

NV
26 tháng 3 2019

\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{101}{3^{101}}\) (1)

\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{100}{3^{101}}+\frac{101}{3^{102}}\) (2)

Trừ (1) cho (2):

\(\frac{2}{3}A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{101}}-\frac{101}{3^{102}}=B-\frac{101}{3^{102}}\)

\(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{101}}\)

\(\Rightarrow\frac{1}{3}B=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{101}}+\frac{1}{3^{102}}\)

\(\Rightarrow\frac{1}{3}B+\frac{1}{3}-\frac{1}{3^{102}}=\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^{101}}=B\)

\(\Rightarrow\frac{2}{3}B=\frac{1}{3}-\frac{1}{3^{102}}\Rightarrow B=\frac{1}{2}\left(1-\frac{1}{3^{101}}\right)=\frac{1}{2}-\frac{1}{2.3^{101}}\Rightarrow B< \frac{1}{2}\)

\(\Rightarrow A=\frac{3}{2}\left(B-\frac{101}{3^{102}}\right)< \frac{3}{2}B< \frac{3}{2}.\frac{1}{2}=\frac{3}{4}\)

28 tháng 2 2020

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

\(\Leftrightarrow3A=3^2+3^3+3^4+3^5+....+3^{101}\)

\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)

\(\Leftrightarrow2A=3^{101}-3\)

\(\Leftrightarrow A=\frac{3^{101}-3}{2}< 3^{100}-1\)

\(\Leftrightarrow A< B\)

28 tháng 2 2020

a. tính A = 3+3^2+3^3+3^4+.....+3^100

3A=3^2+3^3+3^4+3^5+....+3^100

3A-A=(3^2+3^3+3^4+....+3^101)-(3+3^2+3^3+3^4+.....+3^100)=3^101-3=3^100

mà B=3^100-1 => A<B

18 tháng 1 2016

troi,zậy mà khó

 

25 tháng 9 2016

ai trả lời giúp tui đi