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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(S=1+2+2^2+2^3+...+2^{2022}=\dfrac{2^{2022+1}-1}{2-1}=2^{2023}-1\)
b) \(S=1+4+4^2+4^3+...+4^{2022}=\dfrac{4^{2022+1}-1}{4-1}=\dfrac{4^{2023}-1}{3}\)
\(S=1+2+2^2+2^3+...+2^{2022}\\ 2S=2+2^2+2^3+2^4+...+2^{2023}\\ 2S-S=2+2^2+2^3+2^4+...+2^{2023}-1-2-2^2-2^3-...-2^{2022}\\ S=2^{2023}-1\\ S=4+4^2+4^3+...+4^{2022}\\ 4S=4^2+4^3+4^4+...+4^{2023}\\ 4S-S=4^2+4^3+4^4+...+4^{2023}-4-4^2-4^3-...-4^{2023}\\ 3S=4^{2023}-4\\ S=\dfrac{4^{2023}-4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1 + 4 + 42 + 43 + ... + 42021
A = 40 + 41 + 42 + 43 +...+ 42021
Xét dãy số 0; 1; 2; 3;...; 2021
Dãy số trên có số số hạng là:
(2021 - 0) : 1 + 1 = 2022
Vậy A có 2022 số hạng
vì 2022 : 3 = 674
Vậy ta nhóm 3 số hạng liên tiếp của A thành một nhóm thì khi đó
A = (1 + 4 + 42) + (43 + 44 + 45) +...+ (42019 + 42020 + 42021)
A = (1 + 4 + 16) + 43.(1 + 4 + 42) + ... +42019.(1 + 4 + 42)
A = 21 + 43.21 +... + 42019.21
A = 21.(1 + 43 + ... + 42019)
21 ⋮ 21 ⇒ 21.(1 + 43 + ...+ 42019) ⋮ 21 ⇒ A ⋮ 21 (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(A=2+2^2+2^3+2^4+...+2^{100}\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(=3\cdot\left(2+2^3+...+2^{99}\right)⋮3\)
b: Ta có: \(B=4+4^2+4^3+...+4^{2022}\)
\(=4\left(1+4\right)+4^3\left(1+4\right)+...+4^{2021}\left(1+4\right)\)
\(=5\cdot\left(4+4^3+...+4^{2021}\right)⋮5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có M ⋮ 25 vì 75 ⋮ 25
Lại có M = 75 ( 42021 + 42020 + ... + 42 + 4 + 1 )
= 75 . 4 ( 22020 + 22019 + ... + 4 + 1 + 0,25 ) ⋮ 4 vì 4 ⋮ 4
Mà ( 25; 4 ) = 1 ⇒ M ⋮ 100
Vậy M ⋮ 100
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=1+4+4^2+...+4^{11}\)
\(\Rightarrow B=\left(1+4\right)+\left(4^2+4^3\right)+...+\left(4^{10}+4^{11}\right)\)
\(\Rightarrow B=\left(1+4\right)+4^2\left(1+4\right)+...+4^{10}\left(1+4\right)\)
\(\Rightarrow B=\left(1+4\right)\left(4^2+...+4^{10}\right)\)
\(\Rightarrow B=5\left(4^2+...+4^{10}\right)⋮5\)
\(B=1+4+4^2+...+4^{11}\)
\(\Rightarrow B=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)...+\left(4^9+4^{10}+4^{11}\right)\)
\(\Rightarrow B=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)...+4^9\left(1+4+4^2\right)\)
\(\Rightarrow B=\left(1+4+4^2\right)\left(4^3+...+4^9\right)\)
\(\Rightarrow B=21\left(4^3+...+4^9\right)⋮21\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=75.4\left(4^{2020}+4^{2019}+...+4+1\right)+75+25=\)
\(=300.\left(4^{2020}+4^{2019}+...+4+1\right)+100=\)
\(=100\left[3.\left(4^{2020}+4^{2019}+...+4+1\right)+1\right]⋮100\)
S = \(\dfrac{1+4+4^2+4^3+4^4+...+4^{2021}}{1-4^{2022}}\)
Đặt tử số là A thì S = \(\dfrac{A}{1-4^{2022}}\)
A = 1 + 4 + 42 + 43 + 44 + ... + 42021
4A= 4 + 42 + 43 + 44 + 45 + ....+ 42022
4A - A = 4 + 42+43+44+45+...+42022- (1+4+42+43+44+...+42021)
4A = 4 + 42 + 43 + 44+45+42022 - 1 - 4 - 42 - 43 - 44 - ... - 42021
3A = (4 - 4) +(42 - 42) + (43-43) + (44 -44) +...+(42021- 42021)+42022- 1
3A = 42022 - 1
A = \(\dfrac{4^{2022}-1}{3}\)
S = \(\dfrac{4^{2022}-1}{3}\). \(\dfrac{1}{1-4^{2022}}\)
S = - \(\dfrac{1}{3}\)
Ta đặt: \(A=1+4+4^2+...+4^{2021}\)
\(4A=4+4^2+4^3+...+4^{2022}\)
\(4A-A=4+4^2+4^3+...+4^{2022}-1-4-4^2-...-4^{2021}\)
\(3A=4^{2022}-1\)
\(A=\dfrac{4^{2022}-1}{3}\)
\(\Rightarrow S=\dfrac{1+4+4^2+...+4^{2021}}{1-4^{2022}}\)
\(=\dfrac{\left(4^{2022}-1\right):3}{1-4^{2022}}\)
\(=\dfrac{\left(4^{2022}-1\right)\cdot\dfrac{1}{3}}{-\left(4^{2022}-1\right)}\)
\(=-\dfrac{1}{3}\)