Cho A = 1/2+1/mũ 2+1/ mũ 3 +…+1/2 mũ 2015+1/ 2 mũ 2016 Chứng tỏ A<1 Nhanh mình cần gấp mai thi rồi 11/08/2021 Bởi Aaliyah Cho A = 1/2+1/mũ 2+1/ mũ 3 +…+1/2 mũ 2015+1/ 2 mũ 2016 Chứng tỏ A<1 Nhanh mình cần gấp mai thi rồi
`A=1/2 +1/2^2 +1/2^3+…+1/(2^(2016))` `⇔2A=1+1/2+1/2^2+…+1/(2^(2015))` `⇔2A-A=1-1/(2^(2016))` `⇔A=1-1/(2^(2016))<1` Bình luận
`A=1/2+1/(2²)+1/(2³)+…+1/(2^{2015})+1/(2^{2006})` `2A=1+1/2+1/(2^2)+…+1/(2^{2015})` `2A-A=(1+1/2+1/(2²)+…+1/(2^{2015})-(1/2+1/(²)+1/(2³)+…+1/(2^{2015}))+1/(2^{2016}))` `2A-A=1-1/(2^{2016})` Mà `1-1/(2^{2016})<=1` Vậy `A<=1` Bình luận
`A=1/2 +1/2^2 +1/2^3+…+1/(2^(2016))`
`⇔2A=1+1/2+1/2^2+…+1/(2^(2015))`
`⇔2A-A=1-1/(2^(2016))`
`⇔A=1-1/(2^(2016))<1`
`A=1/2+1/(2²)+1/(2³)+…+1/(2^{2015})+1/(2^{2006})`
`2A=1+1/2+1/(2^2)+…+1/(2^{2015})`
`2A-A=(1+1/2+1/(2²)+…+1/(2^{2015})-(1/2+1/(²)+1/(2³)+…+1/(2^{2015}))+1/(2^{2016}))`
`2A-A=1-1/(2^{2016})`
Mà `1-1/(2^{2016})<=1`
Vậy `A<=1`