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Tính \(S=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
Dùng sai phân như sau
\(2S-S=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)=1-\frac{1}{256}\)
Vậy \(S=1-\frac{1}{256}\)
Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
=> 2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
=> 2A - A = (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128) - (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)
=> A = 1 - 1/256
=> A = 255/256
Vậy: ...
\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{128}-\dfrac{1}{256}\right)\)
\(A=1-\dfrac{1}{256}\)
\(A=\dfrac{255}{256}\)
5/11+1/2+2/5+6/11+3/4+16/25+5/16
=1441/400
19/4+37/100+1/8+132/25+5/2+37/12
=1933/120
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
\(a=2.\)\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{..1}{256}\)
\(A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)
\(2A-A=1+\frac{1}{2}+\frac{1}{4}+\frac{....1}{128}-\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)\(+...+\frac{1}{256}\)
\(A=1-\frac{1}{256}\)\(A=\frac{255}{256}\)
Mẫu số chung là 256.
1/2=128/256 ; 1/4=64/256 ; 1/8=32/256 ; 1/16=6/256 ; 1/32=8/256 ; 1/64=4/256 ; 1/128=2/256 .1/256 giữ nguyên .
a=128+64+32+6+8+4+2+1/256
a=245/256
Ta có : \(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}+\frac{1}{2^8}\)
\(\Rightarrow2A=1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^6}+\frac{1}{2^7}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^6}+\frac{1}{2^7}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}+\frac{1}{2^8}\right)\)
\(\Rightarrow A=1-\frac{2}{8}=\frac{256}{256}-\frac{1}{256}=\frac{255}{256}\)
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