Toán A=1+3+3^2+3^3++…+3^11.Chứng tỏ B chia hết cho 13,chia hết cho 40 09/07/2021 By Eliza A=1+3+3^2+3^3++…+3^11.Chứng tỏ B chia hết cho 13,chia hết cho 40
`A=1+3+3^2+3^3+3^4+….+3^11` `->A=(1+3+3^2)+(3^3+3^4+3^5)+…+(3^9+3^10+3^11)` `->A=(1+3+3^2)+3^3(1+3+3^2)+…+3^9(1+3+3^2)` `->A=13.1+3^{3}.13+3^{9}.13` `->A=13(1+3^3+…+3^9)` Vì `13\vdots13` `->13(1+3^3+…+3^9)\vdots13` `->A\vdots13` `A=1+3+3^2+3^3+3^4+….+3^11` `->A=(1+3+3^2+3^3)+(3^4+3^5+3^6+3^7)+(3^8+3^9+3^10+3^11)` `->A=(1+3^2+3^3)+3^4(3+3^2+3^3)+3^8(3+3^2+3^3)` `->A=40.1+3^{4}.40+3^{8}.40` `->A=40(1+3^4+3^8)` Vì `40\vdots40` `->40.1+3^4.40+3^8.40\vdots` `->A\vdots40` Trả lời
Chứng minh $A$ chia hết cho $13$ $A = 1+3 + 3^2 + 3^3 + …. + 3^{11}$ $⇔ A = (1+3+3^2) + (3^3 + 3^4 + 3^5) + …. + (3^9 + 3^{10} + 3^{11})$ $⇔ A = 13 + 3^3.(1+3+3^2) + …. + 3^9.(1+3^1+3^2)$ $⇔ A = 13 + 3^3. 13 + …. + 3^9 . 13$ $⇔ A = 13.(1+3^3 + … + 3^9) \vdots 13$($đ.p.c.m$) Chứng minh $A$ chia hết cho $40$ $A = 1+3 + 3^2 + 3^3 + …. + 3^{11}$ $⇔ A = (1+3+3^2+3^3) + (3^4 + 3^5 + 3^6+3^7) + (3^8 + 3^9 + 3^{10} + 3^{11})$ $⇔ A = 40 + 3^4.(1+3 + 3^2 + 3^3) + 3^8.(1 + 3 + 3^2 + 3^3)$ $⇔ A =40 + 3^4. 40 + …. + 3^8 . 40$ $⇔ A = 40.(1+3^4 + … + 3^8) \vdots 40$($đ.p.c.m$). Trả lời
`A=1+3+3^2+3^3+3^4+….+3^11`
`->A=(1+3+3^2)+(3^3+3^4+3^5)+…+(3^9+3^10+3^11)`
`->A=(1+3+3^2)+3^3(1+3+3^2)+…+3^9(1+3+3^2)`
`->A=13.1+3^{3}.13+3^{9}.13`
`->A=13(1+3^3+…+3^9)`
Vì `13\vdots13`
`->13(1+3^3+…+3^9)\vdots13`
`->A\vdots13`
`A=1+3+3^2+3^3+3^4+….+3^11`
`->A=(1+3+3^2+3^3)+(3^4+3^5+3^6+3^7)+(3^8+3^9+3^10+3^11)`
`->A=(1+3^2+3^3)+3^4(3+3^2+3^3)+3^8(3+3^2+3^3)`
`->A=40.1+3^{4}.40+3^{8}.40`
`->A=40(1+3^4+3^8)`
Vì `40\vdots40`
`->40.1+3^4.40+3^8.40\vdots`
`->A\vdots40`
Chứng minh $A$ chia hết cho $13$
$A = 1+3 + 3^2 + 3^3 + …. + 3^{11}$
$⇔ A = (1+3+3^2) + (3^3 + 3^4 + 3^5) + …. + (3^9 + 3^{10} + 3^{11})$
$⇔ A = 13 + 3^3.(1+3+3^2) + …. + 3^9.(1+3^1+3^2)$
$⇔ A = 13 + 3^3. 13 + …. + 3^9 . 13$
$⇔ A = 13.(1+3^3 + … + 3^9) \vdots 13$($đ.p.c.m$)
Chứng minh $A$ chia hết cho $40$
$A = 1+3 + 3^2 + 3^3 + …. + 3^{11}$
$⇔ A = (1+3+3^2+3^3) + (3^4 + 3^5 + 3^6+3^7) + (3^8 + 3^9 + 3^{10} + 3^{11})$
$⇔ A = 40 + 3^4.(1+3 + 3^2 + 3^3) + 3^8.(1 + 3 + 3^2 + 3^3)$
$⇔ A =40 + 3^4. 40 + …. + 3^8 . 40$
$⇔ A = 40.(1+3^4 + … + 3^8) \vdots 40$($đ.p.c.m$).