Toán 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 18/07/2021 By Kennedy 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 Trả lời
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` ___________________________________________________________________ `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` ________________________________________________________________________ `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` _______________________________________________________ `B=2^7-1` `⇒B<2^7` `⇒A<1` `(đpcm)` Trả lời