Cho A =3+3^3+3^5+3^7+…+3^97 +3^99. Chứng minh rằng A chia hết cho 30 25/11/2021 Bởi Allison Cho A =3+3^3+3^5+3^7+…+3^97 +3^99. Chứng minh rằng A chia hết cho 30
A=3+3^3+3^5+3^7+…+3^97+3^99 A=(3+3^2.3+3^4.3^6.3)+…+(3^93+3^2.3^93+3^4+3^93+3^6.3^93) A=3.(3^2+3^4+3^6)+…+3^93.(3^2+3^4+3^6) A=3.819+…+3^93.819 chia hết 21 . Là nhân nhé Bình luận
A=3+3^3+3^5+3^7+…+3^97+3^99 A=(3+3^2.3+3^4.3^6.3)+…+(3^93+3^2.3^93+3^4+3^93+3^6.3^93) A=3.(3^2+3^4+3^6)+…+3^93.(3^2+3^4+3^6) A=3.819+…+3^93.819 chia hết cho 30 Bình luận
A=3+3^3+3^5+3^7+…+3^97+3^99
A=(3+3^2.3+3^4.3^6.3)+…+(3^93+3^2.3^93+3^4+3^93+3^6.3^93)
A=3.(3^2+3^4+3^6)+…+3^93.(3^2+3^4+3^6)
A=3.819+…+3^93.819 chia hết 21
. Là nhân nhé
A=3+3^3+3^5+3^7+…+3^97+3^99
A=(3+3^2.3+3^4.3^6.3)+…+(3^93+3^2.3^93+3^4+3^93+3^6.3^93)
A=3.(3^2+3^4+3^6)+…+3^93.(3^2+3^4+3^6)
A=3.819+…+3^93.819 chia hết cho 30