PH
0
K
Khách
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KV
1
27 tháng 9 2018
Ta có:
\(3D=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3D-D=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)
\(2D=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
Đặt \(E=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3E=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3E-E=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2E=3-\frac{1}{3^{99}}< 3\)
\(E< \frac{3}{2}\)
\(2D< \frac{3}{2}-\frac{1}{3^{100}}< \frac{3}{2}\)
\(D< \frac{3}{4}\)
Vậy...
19 tháng 8 2016
Theo bài ta có:
\(=\frac{\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{99}{3^{98}}+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\right)}{2}\)
\(=\frac{\left(1-\frac{100}{3^{100}}\right)+\left(\frac{2}{3}-\frac{1}{3}\right)+...+\left(\frac{99}{3^{98}}-\frac{98}{3^{98}}\right)+\left(\frac{100}{3^{99}}-\frac{99}{3^{99}}\right)}{2}\)
\(=\frac{\left(1-\frac{100}{3^{100}}\right)+\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)}{2}< \frac{1+\frac{1}{2}}{2}=\frac{3}{2}:2=\frac{3}{4}\)
Đpcm
NT
2
DA
5 tháng 3 2020
P = 1 + 32 + 34 + 36+......+3100
32 P= 32(1 + 32 + 34 + 36+......+3100)
32P= 32 + 34 + 36+......+3100+3102
32P= (32 + 34 + 36+......+3100+3102)- (1 + 32 + 34 + 36+......+3100 )
32 P= 3102 - 1
P= (3102 -1) :9
Q = (917)3 / 23
Q = 951 / 8
Q = (32)51 /8
Q = 3102 /8
Q= 3102 :8
=> P > Q
Vậy...
K chắc nha b
HD
5 tháng 3 2020
xét P=1+3^2+3^4+3^6+3^8+....+3^100
=> 3^2.P=3^2+3^4+3^6+3^8+3^10+...+3^102
9.P-P=(3^2+3^4+3^6+3^8+3^10+...+3^102)-(1+3^2+3^4+3^6+3^8+....+3^100)
8P=3^102-1
P=\(\frac{3^{102}-1}{8}\)
Xét Q :
\(\left(\frac{9^{17}}{2}\right)^3=\left[\frac{\left(3^2\right)^{17}}{2}\right]^3=\frac{\left(3^{34}\right)^3}{8}=\frac{3^{102}}{8}\)
mà 3^102-1<3^102
=>P<Q
20 tháng 8 2018
a)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(\Rightarrow2A< 1\)
\(\Rightarrow A< \frac{1}{2}\)
C
0
N
23 tháng 7 2015
a)\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(A=1-\frac{1}{2^{50}}<1\)
Vậy \(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}<1\)
b)\(B=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3B-B=2B=1-\frac{1}{3^{100}}\)
\(B=\frac{1-\frac{1}{3^{100}}}{2}\)
Vì \(1-\frac{1}{3^{100}}<1\)nên\(\frac{1-\frac{1}{3^{100}}}{2}<\frac{1}{2}\)
Vậy \(B=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}<\frac{1}{2}\)
c) \(C=\frac{1}{4^1}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{999}}+\frac{1}{4^{1000}}\)
\(4C=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{998}}+\frac{1}{4^{999}}\)
\(4C-C=3C=1-\frac{1}{4^{1000}}\)
\(C=\frac{1-\frac{1}{4^{1000}}}{3}\)
Vì \(1-\frac{1}{4^{1000}}<1\)nên\(\frac{1-\frac{1}{4^{1000}}}{3}<\frac{1}{3}\)
Vậy \(C=\frac{1}{4^1}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{999}}+\frac{1}{4^{1000}}<\frac{1}{3}\)
VH
0