Toán 7 mũ 1=7 mũ2+7 mũ3+…+7 mũ 2016: hết cho 400 14/11/2021 By Melanie 7 mũ 1=7 mũ2+7 mũ3+…+7 mũ 2016: hết cho 400
7+7² +7³ +…+$7^{2016}$ =(7+7²+7³+$7^{4}$) + ($7^{5}$ +$7^{6}$+$7^{7}$+$7^{8}$) +…+ ($7^{2013}$+ $7^{2014}$+$7^{2015}$ + $7^{2016}$) =7(1+7+7²+7³)+ $7^{5}$(1+7+7²+7³) +…+ $7^{2013}$(1+7+7²+7³) =(1+7+7²+7³)(7+$7^{5}$+…+ $7^{2013}$) =400.(7+$7^{5}$+…+ $7^{2013}$) chia hết cho 400 hay 7+7² +7³ +…+$7^{2016}$ chia hết cho 400 Trả lời
`7 ^1 + 7^2 +7^3 +…+7^2016` `= (7 ^1 + 7^2 +7^3 + 7^4)…+(7^2013 + 7^2014 + 7^2015 + 7^2016)` `= 400 . 7^1 + …. + 400 . 7^2013` `= 400 . (7^1 + … + 7^2013)` `=> 7 ^1 + 7^2 +7^3 +…+7^2016` chia hết `400` Trả lời
7+7² +7³ +…+$7^{2016}$
=(7+7²+7³+$7^{4}$) + ($7^{5}$ +$7^{6}$+$7^{7}$+$7^{8}$) +…+ ($7^{2013}$+ $7^{2014}$+$7^{2015}$ + $7^{2016}$)
=7(1+7+7²+7³)+ $7^{5}$(1+7+7²+7³) +…+ $7^{2013}$(1+7+7²+7³)
=(1+7+7²+7³)(7+$7^{5}$+…+ $7^{2013}$)
=400.(7+$7^{5}$+…+ $7^{2013}$) chia hết cho 400
hay 7+7² +7³ +…+$7^{2016}$ chia hết cho 400
`7 ^1 + 7^2 +7^3 +…+7^2016`
`= (7 ^1 + 7^2 +7^3 + 7^4)…+(7^2013 + 7^2014 + 7^2015 + 7^2016)`
`= 400 . 7^1 + …. + 400 . 7^2013`
`= 400 . (7^1 + … + 7^2013)`
`=> 7 ^1 + 7^2 +7^3 +…+7^2016` chia hết `400`