So sánh :
\(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{1000}}\) với \(\frac{1}{3}\)
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IM
27 tháng 11 2016
Ta có :
\(C=\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{1000}}\)
\(\Rightarrow4C=1+\frac{1}{4}+.....+\frac{1}{4^{1999}}\)
\(\Rightarrow4C-C=\left(1+\frac{1}{4}+.....+\frac{1}{4^{1999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{1000}}\right)\)
\(\Rightarrow3C=1-\frac{1}{4^{1000}}\)
\(\Rightarrow C=\frac{1}{3}-\frac{1}{3.4^{1000}}< \frac{1}{3}\)
=> C < 1 / 3
27 tháng 11 2016
Ta có:
\(C=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{1000}}\)
\(\Rightarrow4C=1+\frac{1}{4}+...+\frac{1}{4^{999}}\)
\(\Rightarrow4C-C=\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}+\frac{1}{4^{1000}}\right)\)
\(\Rightarrow3C=1-\frac{1}{4^{1000}}\)
\(\Rightarrow C=\left(1-\frac{1}{4^{1000}}\right).\frac{1}{3}\)
\(\Rightarrow C=\frac{1}{3}-\frac{1}{4^{1000}.3}\)
Mà \(\frac{1}{3}>\frac{1}{3}-\frac{1}{4^{1000}.3}\)
\(\Rightarrow C< \frac{1}{3}\)
Vậy \(C< \frac{1}{3}\)
10 tháng 7 2016
\(M=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{1000}}\)
\(4M=\frac{4}{4}+\frac{4}{4^2}+...+\frac{4}{4^{1000}}\)
\(4M=1+\frac{1}{4}+\frac{1}{4^2}+..+\frac{1}{4^{999}}\)
\(4M-M=\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{1000}}\right)\)
\(3M=1-\frac{1}{4^{1000}}\)
\(M=\left(1-\frac{1}{4^{1000}}\right):3\)
\(M=\frac{4^{1000}-1}{4^{1000}}:3\)
\(M=\frac{4^{1000}-1}{3.4^{1000}}\)
\(\frac{1}{3}=\frac{4^{1000}}{3.4^{1000}}\)
vì \(\frac{4^{1000}-1}{4^{1000}}< \frac{4^{1000}}{3.4^{1000}}\)
nên \(M< \frac{1}{3}\)
15 tháng 1 2019
Sửa đề : \(A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+......+\frac{1}{2^{199}}\)
\(\Rightarrow2A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{198}}\)
\(\Rightarrow2A-A=A=\frac{1}{2}-\frac{1}{2^{199}}< \frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
2 tháng 8 2019
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
3 tháng 8 2018
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1
tương tự nha
3 tháng 8 2018
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}< 1\)
XO
12 tháng 8 2019
B = \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}...+\frac{1}{1+2+3+...+2019}\)
= \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2019\times1010}\)
= \(2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2019\times2020}\right)\)
= \(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{2019\times2020}\right)\)
= \(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
= \(2\times\left(\frac{1}{2}-\frac{1}{2020}\right)\)
\(=2\times\frac{1009}{2020}\)
\(=\frac{1009}{1010}< \frac{1010}{1010}=1\)
\(\Rightarrow B< 1\)
Đặt \(A=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+.....+\frac{1}{4^{1000}}\)
\(=>4A=1+\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{999}}\)
\(=>4A-A=\left(1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{1000}}\right)\)
\(=>3A=1-\frac{1}{4^{1000}}=>A=\frac{1-\frac{1}{4^{1000}}}{3}=\frac{1}{3}-\frac{1}{\frac{4^{1000}}{3}}<\frac{1}{3}\)
Vậy.......................