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\(S=1+2+2^2+2^3+...+2^{2020}+2^{2021}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{2020}+2^{2021}\right)\)
\(=3+2^2\left(1+2\right)+...+2^{2020}\left(1+2\right)\)
\(=3+2^2.3+...+2^{2020}.3⋮3\)
VẬY \(S⋮3\)
Trả lời :...........................................
SCSH: (2021 - 1) : 1 = 2020
Tổng: (2021 + 1) : 2 = 1011
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
k nhé
5.(-8).2.(-3) = ( 5 . 2 ) . [ -8 . ( -3) ] = 10 . 24 = 240
5 . ( - 8 ) . 2 . ( - 3 )
= ( 5 . 2 ) . [ - 8 . ( - 3 ) ]
= 10 . 24
= 240
( 2020 - 7 + 3 ) - ( 7 + 3 - 2020 )
= 2020 - 7 + 3 - 7 - 3 + 2020
= ( 2020 + 2020 ) + ( 3 - 3 ) - ( 7 + 7 )
= 4040 + 0 - 14
= 4026
4.( - 5 )2 + 2. ( - 5 ) - 20
= 4. ( - 5 ) . ( - 5 ) + 2 . ( - 5 ) - 20
= ( - 5 ) . [ 4 . ( - 5 ) + 2 ] - 20
= ( - 5 ) . ( - 18 ) - 20
= 90 - 20 = 70
S = 1 - 2 + 22 - 23 + ....... + 22020
2S = 2(1 - 2 + 22 - 23 + ....... + 22020)
2S = 2 - 22 + 23 - 24 + ....... + 22021
S = (2 - 22 + 23 - 24 + ....... + 22021) - (1 - 2 + 22 - 23 + ....... + 22020)
S = 22021 - 1
3S = 3(22021 - 1)
3S - 22021 = 3(22021 - 1) - 22021
3S - 22021 = 3.22021 - 3 - 22021
➤ 3S - 22021 = 22021 . 2 - 3
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 ✅
\(x^{2020}=x\Leftrightarrow x^{2020}-x=0\Leftrightarrow x\left(x^{2019}-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(1+2+2^2+2^3+....+2^{2019}+2^{2020}\)
\(A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+....+\left(2^{2016}+2^{2017}+2^{2018}\right)+2^{2019}+2^{2020}\)
\(A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+.....+2^{2016}\left(1+2+2^2\right)+2^{2019}+2^{2020}\)
\(A=7+2^3.7+2^6.7+2^9.7+....+2^{2016}.7+2^{2019}+2^{2020}\)
\(\text{Ta có:}2^{2019}+2^{2020}=8^{673}+8^{673}.2\equiv1+1.2\left(\text{mod 7}\right)\equiv3\left(\text{mod 7}\right)\Rightarrow A\text{ chia 7 dư 3}\)