Toán tính A=(1-1/22) . (1-1/32) . ……….. . (1-1/20202) 25/11/2021 By Anna tính A=(1-1/22) . (1-1/32) . ……….. . (1-1/20202)
$A = (1-\dfrac{1}{2^2}).(1-\dfrac{1}{3^2})….(1-\dfrac{1}{2020^2})$ Các số trên có dạng $1-\dfrac{1}{a^2} = \dfrac{a^2-1}{a^2} = \dfrac{(a-1).(a+1)}{a^2}$ Do đó $A = (1-\dfrac{1}{2^2}).(1-\dfrac{1}{3^2})….(1-\dfrac{1}{2020^2})$ $ = \dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}….\dfrac{2019.2021}{2020.2020}$ $ = \dfrac{(1.2.3…2019).(3.4…2021)}{2^2.3^2.4^2….2020^2}$ $ = \dfrac{2021}{2020.2}$ $ =\dfrac{2021}{4040}$ Trả lời
$A = (1-\dfrac{1}{2^2}).(1-\dfrac{1}{3^2})….(1-\dfrac{1}{2020^2})$
Các số trên có dạng
$1-\dfrac{1}{a^2} = \dfrac{a^2-1}{a^2} = \dfrac{(a-1).(a+1)}{a^2}$
Do đó $A = (1-\dfrac{1}{2^2}).(1-\dfrac{1}{3^2})….(1-\dfrac{1}{2020^2})$
$ = \dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}….\dfrac{2019.2021}{2020.2020}$
$ = \dfrac{(1.2.3…2019).(3.4…2021)}{2^2.3^2.4^2….2020^2}$
$ = \dfrac{2021}{2020.2}$
$ =\dfrac{2021}{4040}$