![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
`3A=3+3^3+3^3+...+3^{53}`
`3A-A=(3+3^3+3^3+...+3^{53})-(1+3+3^3+3^3+...+3^{52})`
`2A=3^{53}-1`
`A=(3^{53}-1)/2`
b,
`A=1+3+3^3+3^3+...+3^{52}`
`A=(1+3+3^2)+(3^3+3^4+3^5)+....+(3^{50}+3^{51}+3^{52})`
`A=(1+3+3^2)+3^3*(1+3+3^2)+....+3^{50}*(1+3+3^2)`
`A=(1+3+3^2)*(1+3^3+....+3^{50})`
`A=13*(1+3^3+....+3^{50})`
Do `13 \vdots 13 => A=13*(1+3^3+....+3^{50})\vdots 13 `
Vậy `A \vdots 13 `
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 3A = 3.(1+3+32+33+...+399+3100)
3A = 3+32+33+...+3100+3101
Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)
2A = 3101−1
⇒ A = 3101−1
2
Vậy A = 3101−1
2
![](https://rs.olm.vn/images/avt/0.png?1311)
Tham khảo
Ta có: 3A = 3.(1+3+32+33+...+399+3100)(1+3+32+33+...+399+3100)
3A = 3+32+33+...+3100+31013+32+33+...+3100+3101
Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)(3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)
2A = 3101−13101−1
⇒⇒ A = 3101−123101−12
Vậy A = 3101−12
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1-3+3^2-3^3+3^4-...-3^{98}-3^{99}+3^{100}\\ 3A=3-3^2+3^3-3^4-...-3^{98}+3^{99}-3^{100}+3^{101}\\ 3A-A=3^{101}-1\\ \Rightarrow A=\dfrac{3^{101}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=1+3+3^2+3^3+...+3^{2021}$
$3A=3+3^2+3^3+...+3^{2022}$
$\Rightarrow 3A-A=(3+3^2+3^3+...+3^{2022}) - (1+3+3^2+3^3+...+3^{2021})$
$\Rightarrow 2A=3^{2022}-1$
$\Rightarrow A=\frac{3^{2022}-1}{2}$
$B-A=\frac{3^{2022}}{2}-\frac{3^{2022}-1}{2}=\frac{1}{2}$
\(A=1+3+3^2+...+3^{50}\)
\(3A=3+3^2+3^3+...+3^{51}\)
\(3A-A=\left(3+3^2+3^3+...+3^{51}\right)-\left(1+3+3^2+...+3^{50}\right)\)
\(2A=3^{51}-1\)
\(A=\dfrac{3^{51}-1}{2}\)