Đáp án + Giải thích các bước giải:
`a)` Ta có:
`A=1+2+2^{2}+2^{3}+...+2^{2019}`
`→A=(1+2)+(2^{2}+2^{3})+...+(2^{2018}+2^{2019})`
`→A=2^{0}(1+2)+2^{2}(1+2)+....+2^{2018}(1+2)`
`→A=2^{0}.3+2^{2}.3+...+2^{2018}.3`
`→A=3.(2^{0}+2^{2}+...+2^{2018})` $\vdots$ `3`
`---------------`
`A=1+2+2^{2}+2^{3}+...+2^{2019}`
`→A=(1+2^{2})+(2+2^{3})+....+(2^{2017}+2^{2019})`
`->A=2^{0}(1+2^{2})+2^{1}(1+2^{2})+....+2^{2017}(1+2^{2})`
`→A=2^{0}.5+2^{1}.5+....+2^{2017}.5`
`→A=5.(2^{0}+2^{1}+....+2^{2017})` $\vdots$ `5`
`---------------`
`A=1+2+2^{2}+2^{3}+...+2^{2019}`
`→A=(1+2+2^{2})+(2^{3}+2^{4}+2^{5})+...+(2^{2017}+2^{2018}+2^{2019})`
`→A=2^{0}(1+2+2^{2})+2^{3}(1+2+2^{2})+...+2^{2017}(1+2+2^{2})`
`→A=2^{0}.7+2^{3}.7+...+2^{2017}.7`
`→A=7.(2^{0}+2^{3}+....+2^{2017})` $\vdots$ `7`
`---------------`
`A=1+2+2^{2}+2^{3}+...+2^{2019}`
`→A=(2+2^{2}+2^{3}+2^{4})+...+(2^{2016}+2^{2017}+2^{2018}+2^{2019})+1`
`→A=2^{0}(2+2^{2}+2^{3}+2^{4})+...+2^{2015}(2+2^{2}+2^{3}+2^{4})+1`
`→A=2^{0}.30+...+2^{2015}.30+1`
`→A=30.(2^{0}+...+2^{2015})+1` chia `30` dư `1`
Bạn xem lại đề câu này nha !!!
`b)A=1+2+2^{2}+2^{3}+...+2^{2019}`
`→A+1=2+2+2^{2}+2^{3}+...+2^{2019}`
`→A+1=2^{2}+2^{2}+2^{3}+...+2^{2019}`
`→A+1=2^{3}+2^{3}+...+2^{2019}`
`→A+1=2^{4}+...+2^{2019}`
`→A+1=.............`
`→A+1=2^{2020}`
`→A+1=(2^{1010})^{2}`
Vậy `A+1` là bình phương của một số
