Cho S=1+2+2^2+...+2^2005
Hãy so sánh S với 5*2^2004
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mik bt lm câu 1 thôi nha, bn thông cảm:
a = 2007.2009 b = 20082
=(2008 - 1)(2008 + 1)
= 20082 - 1
Ta có, a = 20082 - 1, b = 20082
mà 20082 - 1 < 20082
=> a < b
\(S=1+2+2^2+2^3+...+2^9\)
Đặt \(2S=2+2^2+2^3+2^4+...+2^{10}\)
\(2S-S=2^{10}-1\) hay \(S=2^{10}-1< 2^{10}\)
\(\Rightarrow\) \(2^{10}=2^2.2^8< 5.2^8\)
Vậy \(S< 5.2^8\)
\(#Tuyết\)
2S=2+2^2+...+2^10
=>S=2^10-1=1023
5*2^8=256*5=1280
=>S<5*2^8
\(S=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2005}}\)
\(2.S=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)
\(2.S-S=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\)
\(S=2-\dfrac{1}{2^{2006}}\)
ta thấy \(\frac{1}{20}\)<\(\frac{1}{3}\)
thì \(\frac{1}{20}\)+...+\(\frac{1}{29}\)<\(\frac{1}{20}\)+...+\(\frac{1}{20}\)<\(\frac{1}{3}\)
vậy \(\frac{1}{20}\)+...+\(\frac{1}{29}\)<\(\frac{1}{3}\)
Sửa đề: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
Ta có: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
\(=\dfrac{1}{20}+\left(\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{30}\right)+\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)\)
\(\Leftrightarrow S>\dfrac{1}{20}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{4}\)
\(\Leftrightarrow S>\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)(đpcm)
Có : \(S=1+2+2^2+2^3+....+2^{99}\)
\(\Rightarrow2S=2+2^2+2^3+....+2^{100}\)
\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+....+2^{99}\right)\)
\(\Rightarrow S=2^{100}-1< 2^{100}\)
Vậy \(S< 2^{100}\)
S=1+2+22+23+....+299
⇒2S=2+22+23+....+2100
⇒2S−S=2100-1
S=2100-1
vì 2100 -1<2100
⇒S<2100
\(S=1+2+2^2+...+2^{2005}\\ 2S=2+2^2+...+2^{2006}\\ 2S-S=S=2^{2006}-1< 2^{2006}+2^{2004}=2^2\cdot2^{2004}+2^{2004}=5\cdot2^{2004}\)
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