Cho A = 3^1+3^2+3^3+….+3^2018+3^2019. Tìm số dư nếu A chia 4 Giúp mình với !!! CẢM ƠN BẸN NHA :))

Question

Cho A = 3^1+3^2+3^3+….+3^2018+3^2019. Tìm số dư nếu A chia 4
Giúp mình với !!!
CẢM ƠN BẸN NHA :))

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Madeline 1 năm 2021-11-30T14:20:12+00:00 2 Answers 1 views 0

Answers ( )

    0
    2021-11-30T14:21:18+00:00

    A=3^1+3^2+3^3=…+3^2018+3^2019

    A=3^1+3^2+3^3+(3^4+3^5+3^6+3^7)+(3^8+3^9+3^10+3^11)…+(3^2016+3^2017+3^2018+3^2019)

    A=39+3^4.(1+3+3^2+3^3)+3^8.(1+3+3^2+3^3)+…+3^2016.(1+3+3^2+3^3)

    A=39+3^4.40+3^8.40+…+3^2016.40

    A=40.(3^4+3^8+…+3^2016)+39 :4 dư 3

    0
    2021-11-30T14:21:28+00:00

    Đáp án:

    `A=3^1+3^2+3^3=…+3^2018+3^2019`

    `A=3^1+3^2+3^3+(3^4+3^5+3^6+3^7)+(3^8+3^9+3^10+3^11)…+(3^2016+3^2017+3^2018+3^2019)`

    `A=39+3^4.(1+3+3^2+3^3)+3^8.(1+3+3^2+3^3)+…+3^2016.(1+3+3^2+3^3)`

    `A=39+3^4.40+3^8.40+…+3^2016.40`

    `A=40.(3^4+3^8+…+3^2016)+39 : 4 dư 3`

    Giải thích các bước giải:

     

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