(1/2^2)+(1/3^2)+(1/4^2)+....(1/2011^2). So sanh tong sau voi 1
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Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{2012}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^{2011}}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^{2011}}\right)\)\(-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{2012}}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^{2012}}\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{2012}}}{2}\)
Vì \(1-\frac{1}{3^{2012}}< 1\Rightarrow A< \frac{1}{2}\)
(1+2+3+4)(1+2+3+4)=(1+2+3+4)2
1.1.1+2.2.2+3.3.3+4.4.4
=13+23+33+43
=(1+2+3+4)3
=> (1+2+3+4).(1+2+3+4) < 1.1.1+2.2.2+3.3.3+4.4.4
( 1 + 2 + 3 + 4)( 1 + 2 + 3 + 4) = 10 . 10 = 100
1.1.1 + 2.2.2 + 3.3.3 + 4.4.4 = 1 + 8 + 27 + 64 = 100
VẬy (1 + 2 + 3 + 4)(1 + 2 + 3 + 4) = 1.1.1 + 2.2.2 + 3.3.3 + 4.4.4
Ta có :
\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\)
\(T=1+\frac{3}{1.2^2}+\frac{4}{2.2^2}+\frac{5}{2^2.2^2}+...+\frac{2016}{2^{2013}.2^2}+\frac{2017}{2^{1014}.2^2}\)
\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{4}+\frac{6}{8}+...+\frac{2016}{x}+\frac{2017}{x}\right)\)
\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2013}}+\frac{2017}{2^{2014}}\right)\)
Đến chỗ này chịu!
đật tông này là A
suy ra A<1/1.2+1/2.3+1/3.4+...+1/2010.2011
Ta có: 1/1.2+1/2.3+1/3.4+...+1/2010.2011=1-1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011
=1-1/2011=2010/2011
Vì 2010/2011<1suy ra A<1 hay 1/2^2+1/3^2+...+1/2011^2