A xin lỗi mình viết sai đề để mình viết lại

(0,25 - 3/4) : 2/3 + (-1/2)^2

\(\left(0,25-\frac{3}{4}\right)\div\frac{2}{3}+\left(\frac{-1}{2}\right)^2\)²

\(=\left(\frac{1}{4}-\frac{3}{4}\right)\times\frac{3}{2}+\frac{1}{4}\)

\(=\frac{-1}{2}\times\frac{3}{2}+\frac{1}{4}\)

\(=\frac{-3}{4}+\frac{1}{4}\)

\(=\frac{-2}{4}=\frac{-1}{2}\)

\(S=1+2+2^2+2^3+...+2^{2020}+2^{2021}\)

\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{2020}+2^{2021}\right)\)

\(=3+2^2\left(1+2\right)+...+2^{2020}\left(1+2\right)\)

\(=3+2^2.3+...+2^{2020}.3⋮3\)

     VẬY \(S⋮3\)

Trả lời :...........................................

SCSH: (2021 - 1) : 1 = 2020

Tổng: (2021 + 1) : 2 = 1011

Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

k nhé

\(S=1+2+2^2+2^3+...+2^{2020}+2^{2021}\)

\(\text{Số số hạng của S là 2022 số , chia làm 1011 cặp , mỗi cặp 2 số .}\)

\(\Leftrightarrow S=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{2020}+2^{2021}\right)\)

\(\Leftrightarrow S=3+2^2\left(1+2\right)+...+2^{2020}\left(1+2\right)\)

\(\Leftrightarrow S=3+2^2\times3+...+2^{2020}\times3\)

\(\Leftrightarrow S=3\left(1+2^2+...+2^{2020}\right)\)

\(\Rightarrow S⋮3\left(đpcm\right)\)

đề bài

\(s=1-2+2^2-2^3.....+2^{2020}\)

chúc bạn học tốt!

Nhầm đề hả bạn 

Bài 2:

Ta thấy: 5² > 4.5

6² > 5.6

7² > 6.7

     ....

2017² > 2016.2017

\(\Rightarrow\)\(\frac{1}{5^2}< \frac{1}{4.5}\)

\(\frac{1}{6^2}< \frac{1}{5.6}\)

\(\frac{1}{7^2}< \frac{1}{6.7}\)

....

\(\frac{1}{2017^2}< \frac{1}{2016.2017}\)

Cộng vế với nhau, ta có:

\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2017^2}\) < \(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2016.2017}\)

\(\Rightarrow\)A < \(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(\Rightarrow\)A < \(\frac{1}{4}-\frac{1}{2017}\)

\(\Rightarrow\)A < \(\frac{1}{4}\)( vì \(\frac{1}{2017}>0\))

k giúp mik ✅