giúp mình vói cho A=1+20+20^2+20^3+...+20^2021. tìm số tự nhiên n biết 20^n-4=19.A+1
2 câu trả lời
`A = 1 + 20 + 20^2 + 20^3 + ... + 20^2021`
`=> 20A = 20 + 20^2 + 20^3 + 20^3 + ... + 20^2022`
`=> 20A - A = ( 20 + 20^2 + 20^3 + 20^3 + ... + 20^2022 ) - ( 1 + 20 + 20^2 + 20^3 + ... + 20^2021 )`
`=> 19A = 20^2022 - 1`
Ta có :
`19A + 1 = 20^2021 - 1 + 1 = 20^2021`
Mà `20^(n-4) = 19A + 1 => 20^(n - 4) = 20^2021`
`=> n - 4 = 2021`
`=> n = 2021 + 4`
`=> n = 2025`
Vậy `n =2025`
`#dtkc`
`A = 1 + 20 + 20^2 + 20^3 + ... + 20^2021`
`20A = 20 + 20^2 + 20^3 + 20^4 + ... + 20^2022`
`20A - A = (20 + 20^2 + 20^3 + 20^4 + ... + 20^2022) - (1 + 20 + 20^2 + 20^3 + ... + 20^2021)`
`19A = 20^2022 - 1`
`A = (20^2022 - 1) : 19`
Ta có:
`20^n - 4 = 19 . A + 1`
`=> 20^n - 4 = 19 . (20^2022 - 1) : 19 + 1`
`=> 20^(n - 4) = 20^2022 - 1 + 1`
`=> 20^(n - 4) = 20^2022`
`=> n - 4 = 2022`
`=> n = 2022 + 4`
`=> n = 2026`
Vậy `n = 2026`