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![](https://rs.olm.vn/images/avt/0.png?1311)
Mình làm ngắn gọn nhé.
\(A=1+2+2^2+...+2^{50}\)
\(\Rightarrow2A=2+2^2+...+2^{51}\)
\(\Rightarrow2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}\)
\(\Rightarrow A=2^{51}-1\)
\(B=1+3+...+3^{66}\)
\(3B=3+3^2+...+3^{67}\)
\(2B=3+3^2+...+3^{67}-1-3-...-3^{66}\)
\(2B=3^{67}-1\)
\(B=\frac{3^{67}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+2+2^2+2^3+2^4+...+2^{50}\)
\(2A=2+2^2+2^3+2^4+...+2^{51}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{51}\right)-\left(1+2+2^2+2^3+2^4+...+2^{50}\right)\)
\(A=2^{51}-1=2\cdot2^{50}-1\)
Mà \(2^{51}=2\cdot2^{50}\)
=> A < 251
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : A = 30 + 31 + 32 + 33 + .... + 350
=> 3A = 31 + 32 + 33 + 34 + ... + 351
Khi đó 3A - A = (31 + 32 + 33 + 34 + ... + 351) - (30 + 31 + 32 + 33 + .... + 350)
=> 2A = 351 - 30
=> A = \(\frac{3^{51}-1}{2}\)
Khi đó A = \(\frac{3^{51}-1}{2}=\frac{3^3.3^{48}-1}{2}=\frac{27.\left(3^4\right)^{12}-1}{2}=\frac{27.\left(...1\right)^{12}-1}{2}\)
\(=\frac{\left(...7\right)-1}{2}=\frac{\left(...6\right)}{2}=\left(...3\right)\)
Vậy A tận cùng là 3
![](https://rs.olm.vn/images/avt/0.png?1311)
tinh tong cua 11,10,9,8,7,6,5,..........,-37,-38
-675 nha ban
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2}.\frac{2}{3}.\)\(...\frac{99}{100}=\frac{1.2.....99}{2.3.....100}=\frac{1.\left(2.....99\right)}{\left(2.3.....99\right).100}=\frac{1}{100}\)
A = 1-2+2^2 - 2^3 +...+ 2^50
2A = 2 - 2^2 + 2^3 - 2^4 +...+ 2^51
2A + A = ( 2 - 2^2 + 2^3 - 2^4 +...+ 2^51 )+(1-2+2^2 - 2^3 +...+ 2^50 )
3A = 2^51 + 1
A = 251 + 1/3