Question

\(A=3+3^2+3^3+...+3^{2000}\)

tim chu so tan cung cua A

Answer 1

\(3+3^2+3^3+......+3^{2000}\)

\(\Rightarrow3A=3^2+3^3+3^4+....+3^{2001}\)

\(\Rightarrow3A-A=\left(3^2+3^3+3^4+....+3^{2001}\right)-\left(3+3^2+3^3+3^4+.....+3^{2000}\right)\)

\(\Rightarrow2A=2^{2001}-3\)

\(\Rightarrow A=\frac{2^{2001}-3}{2}\)

Vậy chữa số tận cùng của A là : 0

Answer 2

3A = 3² + 3³ + ...................+ 3²⁰⁰¹

3A - A = 3²⁰⁰¹ - 3

2A = 3²⁰⁰⁰ .3 - 3 

2A = ....1 .3 - 3

2A = .....3 - 3 

A = ........0 : 2

2A= .......0