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1+2+4+8+16+32+64+128+256+512+1024+2048
=1+(2+8)+(4+16)+(32+128)+(64+256)+(512+2048)+1024
=1+10+20+160+320+2560+1024
=4095
1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 = 4095
k nha công chúa nụ cười =_= ^_^
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{512}-\frac{1}{1024}\)
\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^9}-\frac{1}{2^{10}}\)
\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...+\frac{1}{2^8}-\frac{1}{2^9}\)
\(3A=1-\frac{1}{2^{10}}< 1\)
\(\Rightarrow A< \frac{1}{3}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}=\frac{1023}{1024}\)
BẤM ĐÚNG NHÉ
Đặt \(A=\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(A=\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8}\)
\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)
\(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\right)-\left(\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8}\right)\)
\(A=\frac{1}{2^2}-\frac{1}{2^8}\)
\(A=\frac{1}{4}-\frac{1}{256}=\frac{63}{256}\)
\(\Rightarrow\frac{63}{256}.x=\frac{1}{512}=\frac{1}{2^9}\)
\(\Rightarrow\frac{63}{2^8}.x=\frac{1}{2^9}\)
\(\Rightarrow x=\frac{1}{2^9}:\frac{63}{2^8}=\frac{1}{2^9}.\frac{2^8}{63}=\frac{1}{2.63}=\frac{1}{126}\)
Ủng hộ mk nha !!! ^_^
\(1)C=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{162}\)
\(3C=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{54}\)
\(3C-C=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{54}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{162}\right)\)
\(2C=1-\dfrac{1}{162}\)
\(2C=\dfrac{161}{162}\)
\(C=\dfrac{161}{162}.\dfrac{1}{2}\)
\(C=\dfrac{161}{324}\)
\(2)A=\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}+\dfrac{1}{512}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}\right)-\left(\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{128}+\dfrac{1}{512}\right)\)
\(A=1-\dfrac{1}{512}=\dfrac{511}{512}\)
#)Giải :
\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Lời giải
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
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
Đề bài: Tính
\(A=\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\frac{1}{512}+\frac{1}{2048}\)
\(A=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}+\frac{1}{2^{11}}\)
\(2^2.A=2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\)
\(4A-A=\left(2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}+\frac{1}{2^{11}}\right)\)
\(3A=2-\frac{1}{2^{11}}\)
\(\Rightarrow A=\frac{2-\frac{1}{2^{11}}}{3}\)
Vậy \(A=\frac{2-\frac{1}{2^{11}}}{3}\).
ta có
A= 1/2+ 1/8+1/32+1/128+1/512+1/2048
=> A= 1/2 +1/ 2^3 +1/2^5 +1/2^7+1/2^9+1/2^11
=> 2^2 A=2+1/2+1/2^3+1/2^5+1/2^7+1/2^9
=> 2^2A-A= (2+1/2+1/2^3+1/2^5+1/2^7+1/2^9)-(1/2+1/2^3+/2^5+1/2^7+1/2^9+1/2^11)
=> 3A= 2- 1/2^11
=>3A= 4095/2048
=> A= 1365/2048