A=1/2+1/2^2+1/2^3+…+1/2^2015+1/2^2016 CTR A<1 12/10/2021 Bởi Rylee A=1/2+1/2^2+1/2^3+…+1/2^2015+1/2^2016 CTR A<1
Ta có $A = \dfrac{1}{2} + \dfrac{1}{2^2} + \cdots + \dfrac{1}{2^{2016}}$ Suy ra $\dfrac{1}{2} A = \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{2017}}$ Khi đó $A – \dfrac{1}{2} A = \left( \dfrac{1}{2} + \dfrac{1}{2^2} + \cdots + \dfrac{1}{2^{2016}} \right) – \left( \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{2017}} \right)$ Suy ra $\dfrac{1}{2} A = \dfrac{1}{2} – \dfrac{1}{2^{2017}}$ hay $A = 1 – \dfrac{1}{2^{2016}}< 1$ Vậy $A < 1$ Bình luận
`A=1/2+1/2^2+1/2^3+…+1/2^2015+1/2^2016` `⇒2A=1+1/2+1/2^2+…+1/2^2014+1/2^2015` `⇒2A-A=(1+1/2+1/2^2+…+1/2^2014+1/2^2015)-(1/2+1/2^2+1/2^3+…+1/2^2015+1/2^2016)` `⇒A=1-1/2^2016<1` `⇒A<1` `(đpcm)` Bình luận
Ta có
$A = \dfrac{1}{2} + \dfrac{1}{2^2} + \cdots + \dfrac{1}{2^{2016}}$
Suy ra
$\dfrac{1}{2} A = \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{2017}}$
Khi đó
$A – \dfrac{1}{2} A = \left( \dfrac{1}{2} + \dfrac{1}{2^2} + \cdots + \dfrac{1}{2^{2016}} \right) – \left( \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{2017}} \right)$
Suy ra
$\dfrac{1}{2} A = \dfrac{1}{2} – \dfrac{1}{2^{2017}}$
hay
$A = 1 – \dfrac{1}{2^{2016}}< 1$
Vậy $A < 1$
`A=1/2+1/2^2+1/2^3+…+1/2^2015+1/2^2016`
`⇒2A=1+1/2+1/2^2+…+1/2^2014+1/2^2015`
`⇒2A-A=(1+1/2+1/2^2+…+1/2^2014+1/2^2015)-(1/2+1/2^2+1/2^3+…+1/2^2015+1/2^2016)`
`⇒A=1-1/2^2016<1`
`⇒A<1` `(đpcm)`