Toán chứng minh 1+3+3^2+3^3+………..3^99 chia hết cho 4 17/07/2021 By Madeline chứng minh 1+3+3^2+3^3+………..3^99 chia hết cho 4
Đặt $A=1+3+3^2+3^3+…+3^{99}$ $A=4+3^2.(1+3)+3^4.(1+3)+…+3^{98}.(1+3)$ $A=4+3^2.4+3^4.4+…+3^{98}.4$ $A=4.(1+3^2+3^4+…+3^{98})$ mà $4n\vdots 4$ $→A\vdots 4$ Trả lời
Đáp án: `A\vdots4` Giải thích các bước giải: `text{Chứng minh}``A=1+3+3^2+3^3+………..3^99\vdots 4``text{Ta có:}``A=(1+3)+(3^2+3^3)+(3^4+3^5)+…+(3^98+3^99)``A=4+(3^2 . 1 + 3^2 . 3)+(3^4 . 1+ 3^4 . 3)+…+(3^98 . 1 + 3^98 . 3)``A=4+3^2.(1+3)+3^4.(1+3)+…+3^98.(1+3)``A=4+3^2 . 4+3^4 . 4+…+3^98 . 4` `A=4.1+3^2 . 4+3^4 . 4+…+3^98 . 4``A=4.(1+3^2+3^4+…+3^98)``text{Vì}` `4\vdots4``=>4.(3^2+3^4+…+3^98)\vdots4``text{Vậy}` `A\vdots4` Trả lời
Đặt $A=1+3+3^2+3^3+…+3^{99}$
$A=4+3^2.(1+3)+3^4.(1+3)+…+3^{98}.(1+3)$
$A=4+3^2.4+3^4.4+…+3^{98}.4$
$A=4.(1+3^2+3^4+…+3^{98})$
mà $4n\vdots 4$
$→A\vdots 4$
Đáp án:
`A\vdots4`
Giải thích các bước giải:
`text{Chứng minh}`
`A=1+3+3^2+3^3+………..3^99\vdots 4`
`text{Ta có:}`
`A=(1+3)+(3^2+3^3)+(3^4+3^5)+…+(3^98+3^99)`
`A=4+(3^2 . 1 + 3^2 . 3)+(3^4 . 1+ 3^4 . 3)+…+(3^98 . 1 + 3^98 . 3)`
`A=4+3^2.(1+3)+3^4.(1+3)+…+3^98.(1+3)`
`A=4+3^2 . 4+3^4 . 4+…+3^98 . 4`
`A=4.1+3^2 . 4+3^4 . 4+…+3^98 . 4`
`A=4.(1+3^2+3^4+…+3^98)`
`text{Vì}` `4\vdots4`
`=>4.(3^2+3^4+…+3^98)\vdots4`
`text{Vậy}` `A\vdots4`