cho A= 3+3^2+3^3+…+3^120 chứng minh rằng A chia hết cho 13 CÁC BN GIÚP MK NHA MK ĐANG CẦN GẤP ! 21/07/2021 Bởi Clara cho A= 3+3^2+3^3+…+3^120 chứng minh rằng A chia hết cho 13 CÁC BN GIÚP MK NHA MK ĐANG CẦN GẤP !
$A=3+3^2+3^3+…+3^{120}$ $⇒A=3.(1+3+3^2)+…+3^{118}.(1+3+3^2)$ $⇒A=3.13+…+3^{118}.13$ $⇒A=13.(3+…+3^{118}$$\vdots$13 Vậy A$\vdots$13 Bình luận
Bạn tham khảo nha: A= 3+3^2+3^3+…+3^120 = (3+3^2+3^3)+(3^4+3^5+3^6))+…+(3^118+3^119+3^120) =3.(1+3+3^2)+3^4.(1+3+3^2)+…+3^118.(1+3+3^2) =3.13+3^4.13+…+3^118.13 =13.(3+3^4+…+3^118) ⋮ 13 =>3+3^2+3^3+…+3^120 ⋮ 13 Vậy A ⋮ 13. Bình luận
$A=3+3^2+3^3+…+3^{120}$
$⇒A=3.(1+3+3^2)+…+3^{118}.(1+3+3^2)$
$⇒A=3.13+…+3^{118}.13$
$⇒A=13.(3+…+3^{118}$$\vdots$13
Vậy A$\vdots$13
Bạn tham khảo nha:
A= 3+3^2+3^3+…+3^120
= (3+3^2+3^3)+(3^4+3^5+3^6))+…+(3^118+3^119+3^120)
=3.(1+3+3^2)+3^4.(1+3+3^2)+…+3^118.(1+3+3^2)
=3.13+3^4.13+…+3^118.13
=13.(3+3^4+…+3^118) ⋮ 13
=>3+3^2+3^3+…+3^120 ⋮ 13
Vậy A ⋮ 13.