Question

Cho A=1+3+3^2+3^3+...+3^2000 Biết 2A=3^n-1 Khi đó n=

Answer 1

3A=3+3²+3³+...........+3²⁰⁰¹

3A-A=(3+3²+3³+.............+3²⁰⁰¹)-(1+3+3²+...........+3²⁰⁰⁰)

3A-A=3²⁰⁰¹-1

=>2A=3²⁰⁰¹-3

=>n=2001

Answer 2

ta có 3a = 3 ( 1+ 3 + 3^2 + 3^3 +........+ 3^2000 ) = 3 + 3^2 + 3^3+.......+ 3^2001

ta cũng có 2a = 3a -a = 3 + 3^2 + 3^3 +.......+ 3^2001 - 1 + 3 + 3^2 + 3^3 +.......+ 3^2000

= 3^2001 - 1.            vậy n= 2001

Answer 3

3A= 3+3^2 + 3^3 + .... + 3^2001

3A-A=3+3^2+3^3+....+3^2001-1-3-3^2-....-3^2000

2A=3^2001-1=3^n-1

=>n=2001