Tính A = \(\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}\)
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Ta có: \(\frac{\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...\frac{1}{2014}+2014}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=
= \(\frac{\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+...+\left(\frac{1}{2014}+1\right)+1+2014}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=
= \(\frac{\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}+2015}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=\(\frac{2015.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+1\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=2015
Đặt phân thức trên là D
=> D=(1+1+1+1+...+1+2013/2+2012/3+...+2/2013+1/2014)/(1/2+1/3+1/4+...+1/2014)
=> D=(1+2013/2+1+2012/3+1+2011/4+...+1+2/2013+1+1/2014+1)/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=(2015/2+2015/3+2015/4+...+2015/2013+2015/2014+1)/(1/2+1/3+1/4+...+1/2014)
=> D=[2015*(1/2+1/3+1/4+1/5+....+1/2014)]/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=2015
\(\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2013}{1}+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4024}{2012}-2012}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2013}{1}-1\right)+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4024}{2012}-1\right)}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2012}}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)}\)
\(=\frac{1}{2012}\)
Ủng hộ mk nha ^_-
tớ cần gấp !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
đặt tử là A
A=1+2+2^2+2^3+...+2^2012
2A=2+2^2+2^3+2^4+...+2^2013
2A-A=2+2^2+2^3+2^4+...+2^2013-1-2-2^2-2^3-...-2^2012
A=2^2013-1
đặt mẫu là B
B=2^2014-2
=2(2^2013-1)
từ đó suy ra A/B=(2^2013-1)/2(2^2013-1)=1/2
\(\Rightarrow A=\frac{\left[2+2^2+2^3+...+2^{2013}\right]-\left[1+2+2^2+...+2^{2012}\right]}{2^{2014}-2}\)
\(\Rightarrow A=\frac{2^{2013}-1}{2^{2014}-2}\)