tính tổng G= 1-3+3mũ2-3mũ3+3mũ4-...-3mũ99+3mũ100
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)31x32x33x........x3100
=31+2+3+4+...+100
=3(100+1)x(100-1+1):2
=3101x100:2
=35050
Bài b mình không biết làm
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3^1+3^2+3^3+3^4+...+3^{199}\)
\(3A=3^2+3^3+3^4+3^5+...+3^{200}\)
\(3A-A=\left(3^2+3^3+3^4+...+3^{200}\right)-\left(3^1+3^2+3^3+...+3^{199}\right)\)
\(2A=3^{200}-3^1\)
\(A=\frac{3^{200}-3}{2}\)
=))
Đặt \(A=3^1+3^2+3^3+...+3^{199}\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{200}\)
Lấy 3A trừ A theo vế ta có :
\(3A-A=\left(3^2+3^3+3^4+..+3^{200}\right)-\left(3^1+3^2+3^3+..+3^{199}\right)\)
\(2A=3^{200}-1\)
\(A=\frac{3^{200}-1}{2}\)
Vậy \(3^1+3^2+3^3+..+3^{199}=\frac{3^{200}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(D=3-3^2+3^3-3^4+...+3^9-3^{10}+3^{11}\)
=> \(3D=3^2-3^3+3^4-3^5+...+3^{10}-3^{11}+3^{12}\)
Cộng vế 2 BT trên ta được:
\(D+3D=\left(3-3^2+...+3^{11}\right)+\left(3^2-3^3+...+3^{12}\right)\)
\(\Leftrightarrow4D=3^{12}+3\)
\(\Rightarrow D=\frac{3^{12}+3}{4}\)
G=1-3+32-33+34-...-399+3100
3G=3-32+33-34+35-....-3100+3101
3G+G=(3-32+33-34+35-....-3100+3101)+(1-3+32-33+34-...-399+3100)
4G = 3101+1
G=\(\frac{3^{101}+1}{4}\)