Đáp án:
Giải thích các bước giải:
Ta có :
`1+2^2+2^4+2^6+.....+2^2020`
`=(1+2^2+2^4)+(2^6+2^8+2^{10})+.....+(2^{2016}+2^{2018}+2^{2020})`
`=2^0(1+2^2+2^4)+2^6(1+2^2+2^4)+.....+2^{2016}(1+2^2+2^4)`
`=2^{0}.21+2^{6}.21+.....+2^{2016}.21`
`=21(2^0+2^6+.....+2^{2016})` $\vdots$ `21`
Vậy `1+2^2+2^4+2^6+.....+2^2020` $\vdots$ `21`