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\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+...+\frac{1}{128}=1-\frac{1}{128}=\frac{127}{128}\)
A=1-1/2+1/2-1/3+1/3-1/4+...+1/64-1/128
A=1-1/128
A=127/128
Vậy A=\(\frac{127}{128}\)
B=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
B=1-1/100
B=99/100
Vậy B=\(\frac{99}{100}\)
Ta có : \(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}+\frac{1}{2^8}\)
\(\Rightarrow2A=1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^6}+\frac{1}{2^7}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^6}+\frac{1}{2^7}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}+\frac{1}{2^8}\right)\)
\(\Rightarrow A=1-\frac{2}{8}=\frac{256}{256}-\frac{1}{256}=\frac{255}{256}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}\)
\(=\frac{126}{128}=\frac{63}{64}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(=\frac{1+1+1+1+1+1+1}{2}\)
\(=\frac{7}{2}\)
Đặt \(T=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(T=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{64}-\frac{1}{128}\right)\)
\(\Rightarrow T=1-\frac{1}{128}=\frac{127}{128}\)
1/2 + 1/4 + 1/8 + 1/16+ 1/32 + 1/64 + 1/128
= 64/ 128 + 32/128 + 16/128 +8/128 + 4/128 +2/128 + 1/128
= ( 64 + 32 + 16 + 8 + 4 + 2 + 1 ) /128
= 127/ 128
= 1 - 1/2 + 1/2 - 1/4 + 1/4 - ............ + 1/64 - 1/128
= 1 - 1/128
= 127/128
k nha bn
a) A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 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
A = 1 - 1/512
A = 511/512
b) B = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
3B = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
3B - B = 1 - 1/729
2B = 728/729
B = 364/729
a) A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 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
A = 1 - 1/512
A = 511/512
b) B = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
3B = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
3B - B = 1 - 1/729
2B = 728/729
B = 364/729
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: ...
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\)
\(=1-\frac{1}{128}\)
\(=\frac{127}{128}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+.........+\frac{1}{64}-\frac{1}{128}=\frac{1}{1}-\frac{1}{128}=\frac{127}{128}\)