1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... = ?
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\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)
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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
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= \(\dfrac{49}{2}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{16}-\dfrac{1}{32}\)
= \(\dfrac{784}{32}-\dfrac{16}{32}-\dfrac{8}{32}-\dfrac{4}{32}-\dfrac{2}{32}-\dfrac{1}{32}\)
= \(\dfrac{753}{32}\)
A = \(\dfrac{49}{2}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{16}\) - \(\dfrac{1}{32}\)
A \(\times\) 2 = 49 - 1 - \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{16}\)
A \(\times\) 2 - A = 48 - \(\dfrac{49}{2}\) + \(\dfrac{1}{32}\)
A = \(\dfrac{1536}{32}\) - \(\dfrac{784}{32}\) + \(\dfrac{1}{32}\)
A = \(\dfrac{753}{32}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
Nếu là đề thế này thì đây :
Đặt S = 1/2 + 1/4 +1/8 + 1/16 + 1/32 + .....
S = 1/2 + 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5 +....
2S = 1 + 1/2 + 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5+...
2S - S = 1 + 1/2 + 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5 +. .. - 1/2 - 1/2^2 - 1/2^3 - 1/2^4 - 1/2^5 - ...
S = 1