1/2+1/4+1/8+...+1/128
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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
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
gọi dãy số 1/2+1/4+1/8+...+1/128 là A
A=1/2+1/4+1/8+...+1/128
A=1/2+1/2^2+1/2^3+...+1/2^7
2A=1/2^2+1/^3+1/2^4+...+2^8
2A-A=A
ta có
1/2^2+1/2^3+1/2^4+...+1/2^8-(1/2+1/2^2+1/2^3+...+1/2^7)
=1/2^8-1/2
=Tự tìm ra nhé
=
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......+\frac{1}{128}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+........+\frac{1}{64}\)
Lấy 2A-A ta có:
2A-A=\(\left(1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)
\(\Rightarrow A=1-\frac{1}{128}\)
\(\Rightarrow A=\frac{127}{128}\)
1/2+1/4+1/8+1/16+....+1/128
=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+...+1/64-1/128
=1-1/128
=127/128
Đặt tổng là A
\(2xA=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{64}+\dfrac{1}{128}\)
\(\Rightarrow A=2xA-A=1-\dfrac{1}{256}=\dfrac{255}{256}\)
`1/2+1/4+1/8+...+1/128`
`=1/2x2+1/4x2+1/8x2+...+1/128x2`
`=1+1/2+1/4+1/8+...+1/64`
`=1+1/2+1/4+1/8+...+1/64-1/2-1/4-1/8-...-1/128`
`=1-1/128`
`=127/128`
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{128}\)
\(=1-\dfrac{1}{128}\)
\(=\dfrac{128}{128}-\dfrac{1}{128}\)
\(=\dfrac{127}{128}\)