a)\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+\(\frac{1}{16}\)+\(\frac{1}{32}\)+\(\frac{1}{64}\)
b) \(\frac{1}{3}\)+\(\frac{1}{9}\)+\(\frac{1}{27}\)+\(\frac{1}{81}\)+\(\frac{1}{243}\)
Giúp mình với ai nhanh minh tick cho
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A) \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
2A= \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
2A-A = \(1-\dfrac{1}{32}\)
A= \(\dfrac{31}{32}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(A=1-\frac{1}{256}\)
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3B-B=\left(1+\frac{1}{3}+...+\frac{1}{243}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}\right)\)
\(2B=1-\frac{1}{729}\)
\(B=\frac{1-\frac{1}{729}}{2}\)
\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2C=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2C-C=\left(1+\frac{1}{2}+...+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)\)
\(C=1-\frac{1}{64}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(\Rightarrow2A=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\frac{2}{32}+\frac{2}{64}+\frac{2}{128}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)
\(\Rightarrow A=1-\frac{1}{128}=\frac{128}{128}-\frac{1}{128}=\frac{127}{128}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{128}\)
\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+.....+\left(\frac{1}{64}-\frac{1}{128}\right)\)
\(=1-\frac{1}{128}=\frac{127}{128}\)
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
=1/2-1/4+1/4-1/8+1/8-....+1/156-1/152
=1/2-1/152
=255/512
A=255/512
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
A=14 +18 +116 +132 +164 +1128 +1256 +1512
=12 −14 +14 −18 +....+1256 −1512
=12 −1512
=255512
Vậy A=255512
Phạm Long Khánh
b) 1/3+1/3^2+1/3^3+1/3^4+1/3^5 (goi tong bang M)
3M=1+1/3+1/3^2+1/3^3+1/3^4
3M-M=1-1/3^5
2M=242/243
M=242/243*1/2=121/243
^ là gì vậy