Toán So sánh A=2/2+3/2^2+4/2^3+…+2017/21^2016với 3 21/10/2021 By Gabriella So sánh A=2/2+3/2^2+4/2^3+…+2017/21^2016với 3
Tham khảo `A=\frac{2}{2}+\frac{3}{2^2}+…+\frac{2017}{2^{2016}}` `⇒2A=2+\frac{3}{2}+…+\frac{2017}{2^{2015}}` `⇒2A-A=2+\frac{3}{2}+…+\frac{2017}{2^{2015}}-(\frac{2}{2}+\frac{3}{2^2}+…+\frac{2017}{2^{2016}})` `⇒A=2+(\frac{3}{2}-\frac{2}{2})+….+(\frac{2017}{2^{2015}}-\frac{2016}{2^{2015}})-\frac{2017}{2^{2016}}` `⇒A=2+(\frac{1}{2}+….+\frac{1}{2^{2015}})-\frac{2017}{2^{2016}}` Đặt `B=\frac{1}{2}+….+\frac{1}{2^{2015}}` `⇒2B=1+….\frac{1}{2^{2014}}` `⇒2B-B=1+….\frac{1}{2^{2014}}-(\frac{1}{2}+….+\frac{1}{2^{2015}})` `⇒B=1-\frac{1}{2^{2015}}` `⇒A=2+B-\frac{2017}{2^{2016}}` hay `A=2+1-\frac{1}{2^{2015}}-\frac{2017}{2^{2016}}` `⇒A=3-(\frac{1}{2^{2015}}-\frac{2017}{2^{2016}})` Do đó `A<3` Trả lời
Tham khảo
`A=\frac{2}{2}+\frac{3}{2^2}+…+\frac{2017}{2^{2016}}`
`⇒2A=2+\frac{3}{2}+…+\frac{2017}{2^{2015}}`
`⇒2A-A=2+\frac{3}{2}+…+\frac{2017}{2^{2015}}-(\frac{2}{2}+\frac{3}{2^2}+…+\frac{2017}{2^{2016}})`
`⇒A=2+(\frac{3}{2}-\frac{2}{2})+….+(\frac{2017}{2^{2015}}-\frac{2016}{2^{2015}})-\frac{2017}{2^{2016}}`
`⇒A=2+(\frac{1}{2}+….+\frac{1}{2^{2015}})-\frac{2017}{2^{2016}}`
Đặt `B=\frac{1}{2}+….+\frac{1}{2^{2015}}`
`⇒2B=1+….\frac{1}{2^{2014}}`
`⇒2B-B=1+….\frac{1}{2^{2014}}-(\frac{1}{2}+….+\frac{1}{2^{2015}})`
`⇒B=1-\frac{1}{2^{2015}}`
`⇒A=2+B-\frac{2017}{2^{2016}}`
hay `A=2+1-\frac{1}{2^{2015}}-\frac{2017}{2^{2016}}`
`⇒A=3-(\frac{1}{2^{2015}}-\frac{2017}{2^{2016}})`
Do đó `A<3`