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A= 22+22+23+24+..........+250
2A= 23+23+24+25+..........+251
A= 22+22+23+24+..........+250
2A - A= 23 + 251 - 22 - 22
A= 8+251-4 -4
A= 251
a) A = 251
b) A + 3 - 251=251+3-251
A = 3
\(a,2^{700}=\left(2^7\right)^{100}=128^{100}\)
\(5^{300}=\left(5^3\right)^{100}=125^{100}\)
Có \(128^{100}>125^{100}\Rightarrow2^{700}>5^{300}\)
\(b,S=1+2+2^2+...+2^{50}\)
\(\Rightarrow2S=2+2^2+2^3+...+2^{51}\)
\(\Rightarrow2S-S=S=2^{51}-1< 2^{51}\)
a) Ta có :
\(2^{700}=\left(2^7\right)^{100}=128^{100}\)
\(5^{300}=\left(5^3\right)^{100}=125^{100}\)
Vì \(128^{100}>125^{100}\)\(\Rightarrow\)\(2^{700}>5^{300}\)
Vậy \(2^{700}>5^{300}\)
b) \(S=1+2+2^2+...+2^{50}\)
\(\Rightarrow2S=2+2^2+2^3+...+2^{51}\)
\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{51}\right)-\left(1+2+2^2+...+2^{50}\right)\)
\(\Rightarrow S=2^{51}-1< 2^{51}\)
Vậy S < 251
_Chúc bạn học tốt_
A=\(\frac{1}{51}\)+\(\frac{1}{52}\)+......+\(\frac{1}{100}\)
Ta có:\(\frac{1}{51}\)<\(\frac{1}{100}\)
\(\frac{1}{52}\)<\(\frac{1}{100}\)
...................
\(\frac{1}{100}\)=\(\frac{1}{100}\)
\(\Rightarrow\)A=\(\frac{1}{51}+\frac{1}{52}+\).......\(+\frac{1}{100}\)<\(\frac{1}{100}\times50=\frac{1}{2}\)
Vậy A<\(\frac{1}{2}\)
\(A=1.3.5.7...99=\frac{\left(1.3.5.7...99\right)\left(2.4.6...100\right)}{2.4.6...100}=\frac{1.2.3...100}{\left(2.1\right)\left(2.2\right)...\left(2.50\right)}=\frac{\left(1.2.3...50\right)\left(51.52.53....100\right)}{\left(1.2.3...50\right)\left(2.2.2...2\right)}=\frac{51.52.53...100}{2.2...2}=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}=B\)
2A= 2+22+23+ +250+251
-
A = 1+2+22+23+ +250
2A - A = 251 -1
A = B -1
VẬY A<B
2A=22+23+24+...+250+251
=> 2A-A=(22+23+24+...+250+251) -(2+22+23+24+...+250)
<=> A=251-2
=> A=251-2<251
2A=22+23+24+...+250+251
=>2A-A=( 22+23+24+...+250+251)-(2+22+23+24+...+250)
óA=251-2
=>A=251-2<251