chứng minh rằng (1+1/2).(1+1/2^2)………….(1+1/2^2020)<3 các bạn làm ơn giải giúp mình với 24/09/2021 Bởi Arya chứng minh rằng (1+1/2).(1+1/2^2)………….(1+1/2^2020)<3 các bạn làm ơn giải giúp mình với
(1+$\frac{1}{2}$ ).(1+$\frac{1}{2^2}$ )…(1+1/2^2020)=$\frac{2+1}{2}$ .$\frac{2^2+1}{2^2}$…2^2020+1/2^2020 Ở đây ta nhận thấy: $\frac{2+1}{2}$>$\frac{2^2+1}{2^2}$>…>2^2020+1/2^2020 ⇒$\frac{2+1}{2}$>$\frac{2+1}{2}$ .$\frac{2^2+1}{2^2}$…2^2020+1/2^2020 Mà $\frac{2+1}{2}$ <3 ⇒$\frac{2+1}{2}$ .$\frac{2^2+1}{2^2}$…2^2020+1/2^2020<3 Vậy (1+$\frac{1}{2}$ ).(1+$\frac{1}{2^2}$ )…(1+1/2^2020)<3 好好學習 !=^.^=! Bình luận
=(1+1/2).(1+1/2^2)……………(1+1/2^2020)<3 =1.(1/1-1/2+1/2-1/3+1/3-1/4+1/4-…+1/2019-1/2020)<3 =1.(1/1-1/2020)<3 =1.(2020/2020-1/2020)<3 =1.2019/2020<3 =2019/2020<3 Bình luận
(1+$\frac{1}{2}$ ).(1+$\frac{1}{2^2}$ )…(1+1/2^2020)
=$\frac{2+1}{2}$ .$\frac{2^2+1}{2^2}$…2^2020+1/2^2020
Ở đây ta nhận thấy:
$\frac{2+1}{2}$>$\frac{2^2+1}{2^2}$>…>2^2020+1/2^2020
⇒$\frac{2+1}{2}$>$\frac{2+1}{2}$ .$\frac{2^2+1}{2^2}$…2^2020+1/2^2020
Mà $\frac{2+1}{2}$ <3
⇒$\frac{2+1}{2}$ .$\frac{2^2+1}{2^2}$…2^2020+1/2^2020<3
Vậy (1+$\frac{1}{2}$ ).(1+$\frac{1}{2^2}$ )…(1+1/2^2020)<3
好好學習 !=^.^=!
=(1+1/2).(1+1/2^2)……………(1+1/2^2020)<3
=1.(1/1-1/2+1/2-1/3+1/3-1/4+1/4-…+1/2019-1/2020)<3
=1.(1/1-1/2020)<3
=1.(2020/2020-1/2020)<3
=1.2019/2020<3
=2019/2020<3