Chứng minh rằng:
$\frac{1}{2}$ +$\frac{1}{2²}$ + $\frac{1}{2^{3}}$ + $\frac{1}{2^{4}}$ +$\frac{1}{2^{5}}$+ $\frac{1}{2^{6}}$ + $\frac{1}{2^{7}}$ < 1
Trình bày bằng 3 cách
Chứng minh rằng:
$\frac{1}{2}$ +$\frac{1}{2²}$ + $\frac{1}{2^{3}}$ + $\frac{1}{2^{4}}$ +$\frac{1}{2^{5}}$+ $\frac{1}{2^{6}}$ + $\frac{1}{2^{7}}$ < 1
Trình bày bằng 3 cách
C1: Ta có: $\frac{1}{2}$ +$\frac{1}{2²}$ + $\frac{1}{2^{3}}$ + $\frac{1}{2^{4}}$ +$\frac{1}{2^{5}}$+ $\frac{1}{2^{6}}$ + $\frac{1}{2^{7}}$
= $\frac{2^{6}+2^{5}+2^{4}+2^{3}+2^{2}+2+1}{2^{7}}$
= $\frac{64+32+16+8+4+2+1}{128}$
= $\frac{127}{128}$ < 1
Vậy $\frac{1}{2}$ +$\frac{1}{2²}$ + $\frac{1}{2^{3}}$ + $\frac{1}{2^{4}}$ +$\frac{1}{2^{5}}$+ $\frac{1}{2^{6}}$ + $\frac{1}{2^{7}}$ <1
C2: $\frac{1}{2}$ +$\frac{1}{2²}$ + $\frac{1}{2^{3}}$ + $\frac{1}{2^{4}}$ +$\frac{1}{2^{5}}$+ $\frac{1}{2^{6}}$ + $\frac{1}{2^{7}}$
= $\frac{1}{2}$ +$\frac{1}{4}$ + $\frac{1}{8}$ + $\frac{1}{16}$ +$\frac{1}{32}$+ $\frac{1}{64}$ + $\frac{1}{128}$
= 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 – $\frac{1}{128}$
= $\frac{127}{128}$ < 1
Vậy $\frac{1}{2}$ +$\frac{1}{2²}$ + $\frac{1}{2^{3}}$ + $\frac{1}{2^{4}}$ +$\frac{1}{2^{5}}$+ $\frac{1}{2^{6}}$ + $\frac{1}{2^{7}}$ < 1
C3: Đặt A = $\frac{1}{2}$ +$\frac{1}{2²}$ + $\frac{1}{2^{3}}$ + $\frac{1}{2^{4}}$ +$\frac{1}{2^{5}}$+ $\frac{1}{2^{6}}$ + $\frac{1}{2^{7}}$
Ta có: 2A = 2( $\frac{1}{2}$ +$\frac{1}{2²}$ + $\frac{1}{2^{3}}$ + $\frac{1}{2^{4}}$ +$\frac{1}{2^{5}}$+ $\frac{1}{2^{6}}$ + $\frac{1}{2^{7}}$)
= 1 + $\frac{1}{2}$ +$\frac{1}{2^{2}}$ +…+$\frac{1}{2^{6}}$
Suy ra : 2A – A = (1 + $\frac{1}{2}$ +$\frac{1}{2^{2}}$ +…+$\frac{1}{2^{6}}$ ) – ( $\frac{1}{2}$ +$\frac{1}{2²}$ +…+ $\frac{1}{2^{7}}$ )
⇒ A = 1 + $\frac{1}{2}$ +$\frac{1}{2^{2}}$ +…+$\frac{1}{2^{6}}$ – $\frac{1}{2}$ +$\frac{1}{2²}$ +…+ $\frac{1}{2^{7}}$
= 1 – $\frac{1}{27}$ < 1
Vậy A < 1 hay $\frac{1}{2}$ +$\frac{1}{2²}$ + $\frac{1}{2^{3}}$ + $\frac{1}{2^{4}}$ +$\frac{1}{2^{5}}$+ $\frac{1}{2^{6}}$ + $\frac{1}{2^{7}}$ <1
@Kimetsu No Yaiba
Đặt `A=1/2+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+1/2^7`
`C_1:` `A=1/2+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+1/2^7`
`⇒2A=1+1/2+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6`
`-`
`A=` `1/2+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+1/2^7`
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`A=1-1/2^7`
`⇒A<1` `(đpcm)`
`C_2:` `A=1/2+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+1/2^7`
`-`
`⇒1/2A=` `1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+1/2^7+1/2^8`
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`1/2A=1/2-1/2^8`
`⇒1/2A<1/2`
`⇒A<1` `(đpcm)`
`C_3:` `A=1/2+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+1/2^7`
`⇒A=2^6/2^7+2^5/2^7+2^4/2^7+2^3/2^7+2^2/2^7+2/2^7+1/2^7`
`⇒A={1+2+2^2+2^3+2^4+2^5+2^6}/2^7`
Đặt: `B=1+2+2^2+2^3+2^4+2^5+2^6`
`⇒2B=2+2^2+2^3+2^4+2^5+2^6+2^7`
`-`
`B=1+2+2^2+2^3+2^4+2^5+2^6`
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`B=2^7-1`
`⇒B<2^7`
`⇒A<1` `(đpcm)`