Bài $1)$
`B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3^1+1`
`=>3B=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3^1`
`=>B+3B=(3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3^1+1)+(3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3^1)`
`=>4B=(3^{100}-3^{100})+(3^{99}-3^{99})+...+(3^1-3^1)+3^{101}+1`
`=>4B=3^{101}+1`
`=>B={3^{101}+1}/4`
Vậy `B={3^{101}+1}/4`
Bài $2)$
Ta đặt:
`A=1-2^1+2^2-2^3+2^4-...+2^{2020}`
`=>2A=2^1-2^2+2^3-2^4+2^5-...+2^{2021}`
`=>A+2A=(1-2^1+2^2-2^3+2^4-...+2^{2020})+(2^1-2^2+2^3-2^4+2^5-...+2^{2021})`
`=>3A=(2^1-2^1)+(2^2-2^2)+(2^3-2^3)+...+(2^{2020}-2^{2020})+2^{2021}+1`
`=>3A=2^{2021}+1`
`=>A={2^{2021}+1}/3`
Vậy:
`1-2^1+2^2-2^3+2^4-...+2^{2020}={2^{2021}+1}/3`
