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Bài 3:
Ta có:
\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(...\)+\(\frac{1}{2010^2}\)<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{2009.2010}\)
Xét:\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+.....+\(\frac{1}{2009+2010}\)=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)=\(1-\frac{1}{2010}\)<1
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2010^2}< 1\)
\(\)Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< 1\)
A=1+3+32+33+...+32015
=> 3A=3+32+33+...+32016
=> 3A-A=2A=(3+32+33+...+32016)-(1+3+32+33+...+32015)
=32016-1
=>2A+1=32016=(31013)2 là số chính phương.
Ta có:+) \(A=\frac{2^{19}-3}{2^{20}-3}\)
\(2A=\frac{2^{20}-6}{2^{20}-3}=\frac{\left(2^{20}-3\right)-3}{2^{20}-3}\)
\(2A=1-\frac{3}{2^{20}-3}\)
+)\(B=\frac{2^{20}-3}{2^{21}-3}\)
\(2B=\frac{2^{21}-6}{2^{21}-3}=\frac{\left(2^{21}-3\right)-3}{2^{21}-3}\)
\(2B=1-\frac{3}{2^{21}-3}\)
Vì \(2^{20}-3< 2^{21}-3\)nên \(1-\frac{3}{2^{20}-3}< 1-\frac{3}{2^{21}-3}\)
Hok tốt nha^^
a)Ta có: \(31^{11}< 32^{11}=16^{11}\cdot2^{11}\)
\(16^{11}\cdot2^{12}=16^{14}< 17^{14}\)
Lại có \(16^{11}\cdot2^{11}< 16^{11}\cdot2^{12}\)
\(\Rightarrow31^{11}< 16^{11}\cdot2^{11}< 16^{11}\cdot2^{12}< 17^{14}\)
Vậy \(31^{11}< 17^{14}\)
b)\(A=1+3+3^2+3^3+...+3^{2011}\)
\(3A=3\left(1+3+3^2+3^3+...+3^{2011}\right)\)
\(3A=3+3^2+3^3+...+3^{2012}\)
\(3A-A=\left(3+3^2+...+3^{2012}\right)-\left(1+3+...+3^{2011}\right)\)
\(2A=3^{2012}-1\Rightarrow B=2A+1=3^{2012}-1+1=3^{2012}\)
\(\Rightarrow B=\left(3^{1006}\right)^2\) là số chính phương