A= 1+1/2+1/3+....+1/63+1/64 va B =4
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HT
0
NK
1
30 tháng 9 2017
a) \(A=1+2+2^2+...+2^{63}\)
\(\Rightarrow2A=2.\left(1+2+2^2+...+2^{63}\right)\)
\(\Rightarrow2A=2+2^2+...+2^{64}\)
\(\Rightarrow2A-A=2+2^2+...+2^{64}-\left(1+2+2^2+...+2^{63}\right)\)
\(\Rightarrow A=2+2^2+...+2^{64}-1-2-2^2-...-2^{63}\)
\(\Rightarrow A=2^{64}-1\)
Vì \(2^{64}-1=2^{64}-1\Rightarrow A=B\)
b) \(A=3^4+3^5+...+3^{20}\)
\(\Rightarrow3A=3^5+3^6+...+3^{21}\)
\(\Rightarrow3A-A=3^5+3^6+...+3^{21}-3^4-3^5-...-3^{20}\)
\(\Rightarrow2A=3^{21}-3^4\)
\(\Rightarrow A=\frac{3^{21}-3^4}{2}\)
Mà \(B=\frac{3^{21}-3^4}{2}\Rightarrow A=B\)
ND
0
MV
7 tháng 5 2017
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}+\dfrac{1}{64}\\ =\dfrac{1}{2}+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{17}+\dfrac{1}{18}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{33}+\dfrac{1}{34}+...+\dfrac{1}{64}\right)\)
Ta thấy:
\(\dfrac{1}{3}\) lớn hơn \(\dfrac{1}{4}\)
\(\dfrac{1}{5};\dfrac{1}{6};\dfrac{1}{7}\) lớn hơn \(\dfrac{1}{8}\)
\(\dfrac{1}{9};\dfrac{1}{10};...;\dfrac{1}{15}\) lớn hơn \(\dfrac{1}{16}\)
\(\dfrac{1}{17};\dfrac{1}{18};...;\dfrac{1}{31}\) lớn hơn \(\dfrac{1}{32}\)
\(\dfrac{1}{33};\dfrac{1}{34};...;\dfrac{1}{63}\) lớn hơn \(\dfrac{1}{64}\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>\dfrac{1}{2}+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{32}+\dfrac{1}{32}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{64}+\dfrac{1}{64}+...+\dfrac{1}{64}\right)\\ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\\ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>3\)
Vậy \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>3\)(ĐPCM)
A=\(1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+..........+\left(\frac{1}{33}+\frac{1}{34}+........+\frac{1}{64}\right)\)
\(>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+.......+\left(\frac{1}{64}+\frac{1}{64}+......+\frac{1}{64}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(=1+1+1+1\)
\(=4=B\)
\(\Rightarrow A>B\)