Câu 18:
A = 1 + 3 + 3\(^2\) + 3\(^3\) + ... + 3\(^{2021}\)
3A = 3 + 3\(^2\) + 3\(^3\) + 3\(^4\) + ... + 3\(^{2022}\)
3A - A = (3 + 3\(^2\) + 3\(^3\) + 3\(^4\) + ... + 3\(^{2022}\) ) - (1 + 3 + 3\(^2\) + ... + 3\(^{2021}\))
2A = 3 + 3\(^2\) + 3\(^3\) + 3\(^4\) + ... + 3\(^{2021}\) - 1 - 3 - 3\(^2\) - ... - 3\(^{2021}\)
2A = (3 - 3) + (3\(^2\) - 3\(^2\)) + ... + (3\(^{2021}\) - 3\(^{2021}\)) + (3\(^{2022}\) - 1)
2A = 0 + 0 + ... + 0 + (3\(^{2022}\) - 1)
2A = 3\(^{2022}\) - 1
A = \(\frac{3^{2022}-1}{2}\)
2B - A = \(\frac{3^{2022}}{2}\) - \(\frac{3^{2022}-1}{2}\)
2B - A = \(\frac{3^{2022}-3^{2021}+1}{2}\)
2B - A = \(\frac{0+1}{2}\)
2B - A = \(\frac12\)
Câu 19:
Tìm số nguyên \(x\); y \(\in\) Z; \(x-3\) = y(\(x+2\))
Giải:
\(x-3\) = y(\(x+2\))
\(x-3\) = y\(x+2y\)
\(x-yx\) = 3 + 2y
\(x\left(1-y\right)\) = 3 + 2y
\(x=\frac{3+2y}{1-y}\) (1 ≠ y)
\(x\in\) Z ⇔ (3 + 2y) ⋮ (1 -y)
[-2(1 - y) + 5] ⋮ (1 - y)
5 ⋮ (1 - y)
(1 - y) ∈ Ư(5) = {-5; -1; 1; 5}
Lập bảng ta có:
1-y | -5 | -1 | 1 | 5 |
y | 6 | 2 | 0 | -4 |
\(x\)=\(\frac{3+2y}{1-y}\) | -3 | -7 | 3 | -1 |
\(x\); y ∈ Z | tm | tm | tm | tm |
Theo bảng trên ta có: (\(x;y\)) = (-3; 6); (-7; 2); (3; 0); (-1; 4)
Vậy (\(x;y\)) = (-3; 6); (-7; 2); (3; 0); (-1; 4)
Câu 18
\(A=1+3+3^2+3^3+\cdots+3^{2021}\)
\(\implies3A=3+3^2+3^3+3^4+\cdots+3^{2022}\)
\(\rArr2A=3A-A\)
\(=\left(3+3^2+3^3+3^4+\cdots+3^{2022}\right)-\left(1+3+3^2+3^3+\cdots+3^{2021}\right)\)
\(=3^{2022}-1\)
\(\rArr A=\frac{3^{2022}-1}{2}\)
\(\rArr B-A=\frac{3^{2022}}{2}-\frac{3^{2022}-1}{2}\)
\(=\frac{3^{2022}-3^{2022}+1}{2}\)
\(=\frac12\)
