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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
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{6}+\dfrac{1}{32}=\dfrac{48+24+12+16+3}{96}=\dfrac{103}{96}.\)
1/2+1/4+1/8+1/16+1/32+1/64
= 2 x (1/2+1/4+1/8+1/16+1/32+1/64)
= 1 + 1/2+1/4+1/8+1/16+1/32
=> 2A - A = (1+1/2+1/4+1/8+1/16+1/32) - (1/2+1/4+1/8+1/16+1/32+1/64)
=> A = 1 - 1/64
= 63/64
1/2+1/4+1/8+1/32
=1/2 +1/2^2 +1/2^3 +1/2^4
Đặt A=1/2 +1/2^2 +1/2^3 +1/2^4
2A=2(1/2 +1/2^2 +1/2^3 +1/2^4)
2A=1 + 1/2 +1/2^2 +1/2^3
2A-A=(1 + 1/2 +1/2^2 +1/2^3) - (1/2 +1/2^2 +1/2^3 +1/2^4)
A=1 - 1/2^4 =1 - 1/16
=16/16 -1/16 = 15/16
Vậy tổng A=15/16
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)
\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^5}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^6}=1-\dfrac{1}{2^6}\)
\(2xB=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
\(B=2xB-B=1-\dfrac{1}{64}=\dfrac{63}{64}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(=\frac{16}{32}+\frac{8}{32}+\frac{4}{32}+\frac{2}{32}+\frac{1}{32}\)
\(=\frac{31}{32}\)
Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32
2A = 2/2 + 2/4 + 2/8 + 2/16 + 2/32 = 1 + 1/2 + 1/4 + 1/8 + 1/16
2A - A = 1 + 1/2 + 1/4 + 1/8 + 1/16 - 1/2 - 1/4 - ... - 1/32
A = \(1-\frac{1}{32}=\frac{31}{32}\)
31/32
\(\frac{31}{32}\)