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A=1+2+22+23+24
A=20+21+22+23+24
2A=21+22+23+24+25
2A-A=(21+22+23+24+25)-(20-21+23+24)
A=25-1
Vì 25-1=25-1
Nên A=B
Làm tuong tự với các câu sau
S=\(1+2+2^2+2^3+...+2^{2005}\)
2S=\(2+2^2+2^3+2^4...+2^{2006}\)
2S-S=\(\left(2+2^2+2^3+...+2^{2006}\right)-\left(1+2+2^2+2^3+...+2^{2005}\right)\)
S=\(2^{2006}-1< 2^{2006}=2^{2004}.2^2=4.2^{2004}< 5.2^{2004}\)
\(\Rightarrow2^{2006}-1< 5.2^{2004}\)
Vậy \(\text{S}< 5.2^{2004}\)
S=1+2+22+...+22005
2.S=2+2^2+2^3+...+2^{2006}
2.S=2+22+23+...+22006
2S-S=S=\left(2+2^2+..+2^{2006}\right)-\left(1+2+2^2+..+2^{2005}\right)2S−S=S=(2+22+..+22006)−(1+2+22+..+22005)
S=2^{2006}-1S=22006−1
A=5.2^{2004}=\left(4+1\right).2^{2004}=2^2.2^{2004}+2^{2004}=2^{2006}+2^{2004}A=5.22004=(4+1).22004=22.22004+22004=22006+22004
S<A
Bài 1
\(\frac{2017}{2018}+\frac{2018}{2019}\)và \(\left(\frac{2017+2018}{2018+2019}\right)\)mk chữa lại đề luôn đó
Ta tách :
\(\frac{2017}{\left(2018+2019\right)+2018}\)
đến đây ta tách
\(\frac{2017}{2018+2019}< \frac{2017}{2018}\)
vậy....
mấy câu khác tương tự
2) \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{\frac{2}{2003}+\frac{2}{2004}+\frac{2}{2005}}\)
= \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{2.\frac{1}{2003}+2.\frac{1}{2004}+2.\frac{1}{2005}}\)
=\(\frac{1\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}{2.\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}\)
= \(\frac{1}{2}\)
3) \(2013+\left(\frac{2013}{1+2}\right)+\left(\frac{2013}{1+2+3}\right)+...+\left(\frac{2013}{1+2+3+...+2012}\right)\)
= \(2013.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}\right)\)
= \(2013.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{2025078}\right)\)
= \(2013.2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4050156}\right)\)
=\(4026.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)\)
= \(4026.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
= \(4026.\left(1-\frac{1}{2013}\right)\)
= \(4026.\frac{2012}{2013}\)
=\(4024\)
A = 1 + 2 + 22 + 23 + 24 + ... + 22004
2A = 2 + 22 + 23 + 24 + 25 + ... + 22005
2A - A = 22005 - 1
A = 22005 - 1 = B
Ta có: \(S=1+2+2^2+......+2^{2005}\left(1\right)\)
\(\Rightarrow2S=2+2^2+2^3+.....+2^{2006}\left(2\right)\)
Lấy (2)-(1) ta có: \(2S-S=\left(2+2^2+2^3+.......+2^{2006}\right)\)\(-\left(1+2+2^2+......+2^{2005}\right)\)
\(\Rightarrow S=2^{2006}-1\)
\(\Rightarrow S=2^2.2^{2004}-1\)
\(\Rightarrow S=4.2^{2004}-1\Rightarrow S< 5.2^{2004}\)
Ta có : S = 1 + 2 + 22 + ...... + 22015
=> 2S = 2 + 22 + ...... + 22016
=> 2S - S = 22016 - 1
=> S = 22016 - 1
Ta có: 22016 = 4.22014
Mà 4 < 5 nên S < 5.22014
Ta có :
E = 1 + 2 + 22 + 23 + ... + 22004
=> 2E = 2 + 22 + 23 + 24 + ... + 22005
=> 2E - E = ( 2 + 22 + 23 + 24 + ... + 22005 ) - ( 1 + 2 + 22 + 23 + ... + 22004 )
=> E = 22005 - 1 = F
Vậy E = F
2E=2 + 2^2 + 2^3 + 2^4 +....+ 2^2004+2^2005
2E-E=(2 + 2^2 + 2^3 + 2^4 +....+ 2^2004+2^2005)-(1 + 2 + 2^2 + 2^3 + 2^4 +....+ 2^2004)
E=2^2005-1
Vậy E=F(=2^2005-1)