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đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{256}\)
=> A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+....+\frac{1}{2^8}\)
=> 2A=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^7}\)
=> 2A-A=\(\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^8}\right)\)
=> A=\(1-\frac{1}{2^8}\)
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
= 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 + 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 + 1/256 - 1/512
= 1 - 1/512
= 511/512
A=1/2+1/4+1/8.....+1/256+1/512
2A=1+1/2+1/4+1/8...1/256
A=(1+1/2+1/4+1/8...1/256)-(1/2+1/4+1/8.....+1/256+1/512)
A=1-1/512
A=511/512
511/512
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\right)\)
\(A=1-\frac{1}{512}=\frac{511}{512}\)
Đặt: \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(\Rightarrow2A=2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\right)\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(\Rightarrow2A-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)\(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\right)\)
\(\Rightarrow A=1-\frac{1}{512}\)
\(\Rightarrow A=\frac{511}{512}\)
~ rất vui vì giúp đc bn ~
Các số là :
(512 + 1 ) : 1 + 1 = 512 (so )
Tổng trên là :
(512 - 1 ) .512 : 2= 130816
Vậy tổng trên bằng 130816
1 + 2 + 4 + 8 + .......753........ + 256 + 512 = 1.536
Tk mk nha
~~~chúc bn hok tốt~~~
^_^
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}+\frac{1}{512}\)
\(A\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
\(A\cdot2-A=1-\frac{1}{512}\)
\(A=\frac{511}{512}\)
1+2+4+8+......+256+512
Phân tích :
1+1 = 2
2+2 =4
4+4=8
...
256+256 = 512
( Lấy 1 số cộng vs chính nó ! )
Học tốt !
Bài làm
1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512
= ( 2 + 8 ) + ( 4 + 16 ) + ( 128 + 32 ) + ( 64 + 256 ) + ( 512 + 1 )
= 10 + 20 + 160 + 320 + 513
= 1023
512 + 512=1024
Đặt A = 1+2+4+8+…+256+512
=> 2A= 2+4+8+16+…+512+1024
=> 2A-A=1024-1
=> A= 1023
KL….
NHỚ K NHA!!
Ủa ? Làm lộn :)
1+2+4+8+16+32+64+128+256+512
= ( 1 + 512 ) + ( 2+8) + ( 4+16)+(128+32)+(64+256)
= 513 + 10+20+160+320
=1023
Ta đặt:1+2+4+8+......+256+512
=20 + 21 + 22 +...+28 +29
=2(20 + 21 + 22 +...+28 +29)
= 21 + 22 +...+28 +29 +210
do đó:
2 vế trừ cho nhau
=(21 + 22 +...+28 +29 + 210)-(20 + 21 + 22 +...+28 +29)
=210 - 20
=1024-1
=1023