1/4+1/8+1/16+1/32+1/64+1/128
1+1
2+2
4+4
8+8
16+16
32+32
64+64
128+128
1+1=2
2+2=4
4+4=8
8+8=16
16+16=32
32+32=64
64+64=128
128+128=256
tk m nhé
X*[1/2+1/4+1/8+1/16+1/32+1/64+1/128]=127/128
X x (1/2+1/4+1/8+1/16+1/32+1/64+1/128) = 127/128
X x 127/128 = 127/128
X = 127/128 : 127/128
X = 1
1+1=?
2+2=?
4+4=?
8+8=?
16+16=?
32+32=?
64+64=?
128+128=?
1+1=2
2+2=4
4+4=8
8+8=16
16+16=32
32+32=64
64+64=128
128+128=256
1+1=2
2+2=44+4=8
8+8=16
16+16=32
32+32=64
64+64=128
128+128=256
1/4+1/8+1/16+1/32+1/64+1/128.
\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\frac{32+16+8+4+2+1}{128}=\frac{63}{128}\)
Đặt A=1/4+1/8+1/16+1/32+1/64+1/128.
\(A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)
\(2A=\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)
\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
\(A=\frac{1}{2}-\frac{1}{2^7}\)
A=1/4+1/8+1/16+1/32+1/64+1/128
Ax2-A=(1/4+1/8+1/16+1/32+1/64+1/128)x2-(1/4+1/8+1/16+1/32+1/64+1/128)
Ax2-A=(1/2+1/4+1/8+1/16+1/32+1/64)-(1/4+1/8+1/16+1/32+1/64+1/128)
Ax2-A=1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64+1/64-1/128
A=1/2-1/128
A= 63/128.
tính nhanh p/s 1+ 5/4 + 5/8 + 5/16 + 5/32 + 5/64
b) 1/3 +1/9 + 1/27 + 1/81 +...........+ 1/59049
c) 3/2 + 3/8 + 3/32 +3/128 + 3/512
d) 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 giúp mình với
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
C = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
C= \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)
2C = \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
2C-C = \(1-\dfrac{1}{128}\)
C= \(\dfrac{127}{128}\)
S=1/4+1/8+1/16+1/32+1/64+1/128
S=1/4+1/8+1/16+1/32+1/64+1/128
\(S=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)
\(2S=2\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
\(2S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)
\(2S-S=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
\(S=\frac{1}{2}-\frac{1}{2^7}\)
S=(1/2-1/4)+(1/4-1/8)+(1/8-1/16)+(1/16-1/32)+(1/32-1/64)+(1/64-1/128)
S=1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64+1/64-1/128
S=1/2-(1/4-1/4)+(1/8-1/8)+(1/16-1/16)+(1/32-1/32)+(1/64-1/64)-1/128
S=1/2-1/128
S=63/128
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
Đặt A=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A-A=1-\frac{1}{128}\)
\(A=\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\frac{127}{128}\)
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