cho B=1/2+(1/2)2+(1/2)3+(1/2)4+...+(1/2)2014+(1/2)2015
Chung minh rang :B<1
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TT
1
NT
0
KD
1
K
27 tháng 6 2016
Ta thấy:
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
........................
\(\frac{1}{8^2}< \frac{1}{7.8}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}=1-\frac{1}{8}< 1\)
Vậy B < 1
PN
0
20 tháng 10 2016
1/3^2-1/3^4=3^2/3^4-1/3^4=8/3^4
1/3^6-1/3^8=1./3^4.8/3^4=8/3^8
1/3^2014-1/3^2016=8/3^2004
A/8=1/3^4+1/3^8+...+..1/3^2004
A/(8.3^4)=1/3^8+1/3^12+..+1/3^2008
A(1/8-1/(8.3^4)=1/3^4-1/3^2008=(3^2004-1)/3^2008
10.A(1/3^4)=...
10A=(3^2004-1)/3^2004<1
vậy A<1/10=0,1
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+.....+\left(\frac{1}{2}\right)^{2014}+\left(\frac{1}{2}\right)^{2015}\)
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\)
Ta có: \(2B=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2013}}+\frac{1}{2^{2014}}\)
=>\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2013}}+\frac{1}{2^{2014}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\right)\)
=>\(B=1-\frac{1}{2^{2015}}<1\left(đpcm\right)\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2003}}+\frac{1}{2^{2004}}\)
\(B=2B-B=1-\frac{1}{2005}<1\)