Cho S= 4 ^0 + 4^1+ 4 ²+ 4 ³+…..+ 4^ 59 + 4^ 60. Chứng tỏ S – 1 chia hết cho 21. 27/11/2021 Bởi Ayla Cho S= 4 ^0 + 4^1+ 4 ²+ 4 ³+…..+ 4^ 59 + 4^ 60. Chứng tỏ S – 1 chia hết cho 21.
Ta có : S = $4^{0}$ + $4^{1}$ + 4² + 4³ + … + $4^{58}$ +$4^{59}$ +$4^{60}$ ⇔ S = 1 + 4 + 4² + 4³ + $4^{4}$ + $4^{5}$ +$4^{6}$ + … + $4^{58}$ +$4^{59}$ +$4^{60}$ ⇒ S – 1 = (1 + 4 + 4² + 4³ + $4^{4}$ + $4^{5}$ +$4^{6}$ + … + $4^{58}$ +$4^{59}$ +$4^{60}$) – 1 ⇔ S – 1 = 1 + 4 + 4² + 4³ + $4^{4}$ + $4^{5}$ +$4^{6}$ + … + $4^{58}$ +$4^{59}$ +$4^{60}$ – 1 ⇔ S – 1 = 4 + 4² + 4³ + $4^{4}$ + $4^{5}$ +$4^{6}$ + … + $4^{58}$ +$4^{59}$ +$4^{60}$ ⇔ S – 1 = (4 + 4² + 4³) + ($4^{4}$ + $4^{5}$ +$4^{6}$) + … + ($4^{58}$ +$4^{59}$ +$4^{60}$) ⇔ S – 1 = 4(1 + 4 + 4²) + $4^{4}$(1 + 4 + 4²) + … + $4^{58}$(1 +4 +4²) ⇔ S – 1 = 4.21 + $4^{4}$.21 + … + $4^{58}$.21 ⇔ S – 1 = 21(4 + $4^{4}$ + … + $4^{58}$) chia hết cho 21 Vậy S – 1 chia hết cho 21 Bình luận
Ta có : S = $4^{0}$ + $4^{1}$ + 4² + 4³ + … + $4^{58}$ +$4^{59}$ +$4^{60}$
⇔ S = 1 + 4 + 4² + 4³ + $4^{4}$ + $4^{5}$ +$4^{6}$ + … + $4^{58}$ +$4^{59}$ +$4^{60}$
⇒ S – 1 = (1 + 4 + 4² + 4³ + $4^{4}$ + $4^{5}$ +$4^{6}$ + … + $4^{58}$ +$4^{59}$ +$4^{60}$) – 1
⇔ S – 1 = 1 + 4 + 4² + 4³ + $4^{4}$ + $4^{5}$ +$4^{6}$ + … + $4^{58}$ +$4^{59}$ +$4^{60}$ – 1
⇔ S – 1 = 4 + 4² + 4³ + $4^{4}$ + $4^{5}$ +$4^{6}$ + … + $4^{58}$ +$4^{59}$ +$4^{60}$
⇔ S – 1 = (4 + 4² + 4³) + ($4^{4}$ + $4^{5}$ +$4^{6}$) + … + ($4^{58}$ +$4^{59}$ +$4^{60}$)
⇔ S – 1 = 4(1 + 4 + 4²) + $4^{4}$(1 + 4 + 4²) + … + $4^{58}$(1 +4 +4²)
⇔ S – 1 = 4.21 + $4^{4}$.21 + … + $4^{58}$.21
⇔ S – 1 = 21(4 + $4^{4}$ + … + $4^{58}$) chia hết cho 21
Vậy S – 1 chia hết cho 21