Toán a) A = 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9 + …. + 2/2017.2019 18/09/2021 By Sadie a) A = 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9 + …. + 2/2017.2019
Ta có: $A =\dfrac{2}{1.3} + \dfrac{2}{3.5} + \dfrac{2}{5.7} + …. + \dfrac{2}{2017.2019}$ $⇔ A = \dfrac{1}{1} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{5} + \dfrac{1}{5} – \dfrac{1}{7} + ….. + \dfrac{1}{2017} – \dfrac{1}{2019}$ $⇔A = 1 – \dfrac{1}{2019}$ $⇔ A = \dfrac{2018}{2019}$. Trả lời
$\text {Công thức:}$ `2/[n.(n +2)] = (n +2)/(n.(n +2)) – n/(n.(n +2)) = 1/n – 1/(n +2)` $\text {⇒ Áp dụng công thức trên, ta có:}$ `A = 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9 + …. + 2/2017.2019` `= 1 – 1/3 + 1/3 – 1/5 + 1/5 – 1/7 + 1/7 – 1/9 +….+ 1/2017 – 1/2019` `= 1 – 1/2019` `= 2018/2019` Trả lời
Ta có:
$A =\dfrac{2}{1.3} + \dfrac{2}{3.5} + \dfrac{2}{5.7} + …. + \dfrac{2}{2017.2019}$
$⇔ A = \dfrac{1}{1} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{5} + \dfrac{1}{5} – \dfrac{1}{7} + ….. + \dfrac{1}{2017} – \dfrac{1}{2019}$
$⇔A = 1 – \dfrac{1}{2019}$
$⇔ A = \dfrac{2018}{2019}$.
$\text {Công thức:}$ `2/[n.(n +2)] = (n +2)/(n.(n +2)) – n/(n.(n +2)) = 1/n – 1/(n +2)`
$\text {⇒ Áp dụng công thức trên, ta có:}$
`A = 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9 + …. + 2/2017.2019`
`= 1 – 1/3 + 1/3 – 1/5 + 1/5 – 1/7 + 1/7 – 1/9 +….+ 1/2017 – 1/2019`
`= 1 – 1/2019`
`= 2018/2019`