D=5/4 + 5/4^2 +5/4^3 + 5/4^4 +.....+5/4^99 .Chứng tỏ D<5/3
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MV
8 tháng 5 2017
\(A=\dfrac{5}{4}+\dfrac{5}{4^2}+\dfrac{5}{4^3}+...+\dfrac{5}{4^{99}}\\ 4A=5+\dfrac{5}{4}+\dfrac{5}{4^2}+...+\dfrac{5}{4^{98}}\\ 4A-A=\left(5+\dfrac{5}{4}+\dfrac{5}{4^2}+...+\dfrac{5}{4^{98}}\right)-\left(\dfrac{5}{4}+\dfrac{5}{4^2}+\dfrac{5}{4^3}+...+\dfrac{5}{4^{99}}\right)\\ 3A=5-\dfrac{5}{4^{99}}\\ A=\left(5-\dfrac{5}{4^{99}}\right):3\\ A=\dfrac{5}{3}-\dfrac{5}{4^{99}}:3\\ A=\dfrac{5}{3}-\dfrac{5}{4^{99}\cdot3}< \dfrac{5}{3}\)
Vậy \(A< \dfrac{5}{3}\)
30 tháng 8 2020
a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
NM
1
2 tháng 3 2017
\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}\) <4
=\(\frac{71}{20}\) <4
\(\Rightarrow\)3.55 < 4( đpcm)
D=5/4+5/4^2+5/4^3+....+5/4^99
4D=4(5/4+5/4^2+5/4^3+....+5/4^99)
4D=5+5/4+5/4^2+5/4^3+....+5/4^98
Lấy 4D-D=(5+5/4+5/4^2+5/4^3+....+5/4^98)-(5/4+5/4^2+5/4^3+....+5/4^99)
=>3D=5-5/4^99
D=5/3-5/3x4^99<5/3
=>d<5/3(ĐPCM)
`Answer:`
\(D=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+\frac{5}{4^4}+...+\frac{5}{4^{99}}\)
\(\Rightarrow4D=5+\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{98}}\)
\(\Rightarrow4D-D=\left(5+\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{98}}\right)-\left(5+\frac{5}{4^2}+\frac{5}{4^3}+\frac{5}{4^4}+...+\frac{5}{4^{99}}\right)\)
\(\Rightarrow3D=5-\frac{5}{4^{99}}\)
\(\Rightarrow D=\left(5-\frac{5}{4^{99}}\right):3\)
\(\Rightarrow D=\frac{5}{3}-\frac{5}{4^{99}.3}< \frac{5}{3}\)
Vậy `D<5/3`