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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+3^1+3^2+...+3^{2017}\)
\(3A=3+3^2+3^3+...+3^{2018}\)
\(3A-A=\left(3+3^2+3^3+...+3^{2018}\right)-\left(1+3^1+3^2+...+3^{2017}\right)\)
\(2A=3^{2018}-1\)
\(A=\frac{3^{2018}-1}{2}\)
\(\Rightarrow\)\(B-A=\frac{3^{2018}}{2}-\frac{3^{2018}-1}{2}=\frac{3^{2018}-3^{2018}+1}{2}=\frac{1}{2}\)
Vậy \(B-A=\frac{1}{2}\)
Chúc bạn học tốt ~
ta có: A = 1 + 31 + 32 + ...+ 32017
=> 3A = 31 + 32 + 33 + ....+ 32018
=> 3A - A = 32018 - 1
\(\Rightarrow A=\frac{3^{2018}-1}{2}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3^{2018-1}}{2}}{\frac{3^{2018}}{2}}=\frac{\frac{3^{2018}}{2}}{\frac{3^{2018}}{2}}-\frac{1}{\frac{3^{2018}}{2}}=1-\frac{1}{\frac{3^{2018}}{2}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(3m+4n-5p\right)-\left(3m-4n-5p\right)\)
\(\Rightarrow3m+4n-5p-3m+4n+5p=A\)
\(\Rightarrow A=\left(3m-3m\right)+\left(4n+4n\right)-\left(5p-5p\right)\)
\(\Rightarrow A=0+8n+0=8n\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 25 - (-3)3 = 25 - (-27) = 25 + 27 = 49
b) (-1)2018 = 1
c) (-1)2015 = -1
#Học tốt!!!
~NTTH~
Nhầm câu a, cho mik sửa
25 - (-3)3 = 32 - (-27) = 32 + 27 = 59
~Xin lỗi nhiều nha~
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 2 + 22 + 23 + 24 + ... + 29 + 210
A = ( 2 + 22 ) + ( 23 + 24 ) + ... + ( 29 + 210 )
A = ( 1 + 2 ) . 2 + ( 1 + 2 ) . 23 + ... + ( 1 + 2 ) . 29
A = 3 . 2 + 3 . 23 + ... + 3 . 29
A = 3 . ( 2 + 23 + ... + 29 )
=> A chia hết cho 3
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+210\right)=2\left(2^0+2^1\right)+2^3\left(2^0+2^1\right)+... \)
\(2^0=1,2^1=2,2^0+2^1=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Phần b mình chưa nghĩ ra
a) C = 1 . 2 + 2 . 3 + ... + 49 . 50
=> 3C = 1 . 2 . 3 + 2 . 3 . 3 + ... + 49 . 50 . 3
=> 3C = 1 . 2 . ( 3 - 0 ) + 2 . 3 . ( 4 - 1 ) + ... + 49 . 50 . ( 51 - 48 )
=> 3C = 1 . 2 . 3 - 0 . 1 . 2 + 2 . 3 . 4 - 1 . 2 . 3 + ... + 49 . 50 . 51 - 48 . 49 . 50
=> 3C = 49 . 50 . 51
=> C = 49 . 50 . 17 = 41650
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(A=2+2^2+2^3+...+2^{10}=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)
\(A=2.3+2^3.3+...+2^9.3=3\left(2+2^3+...+2^9\right)\)
=>A chia hết cho 3
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 2 + 22 + 23 + 24 + ... + 210
A = 21 . (1 + 2) + 23 . (1 + 2) + ... + 29 + (1 + 2 )
A = 21 . 3 + 23 . 3 + ... + 29 . 3
A = 3 . (21 + 23 + ... + 29)
Vậy A chia hết cho 3
A=2+2^2+...+2^10=2(1+2)+2^3(1+2)+...+2^9(1+2)=2*3+2^3*3+2^9*3=(2+2^3+...+2^9)*3=> CHIA HẾT CHO 3