Question

A= 1+3+3²+3³+3⁴+.....+3²⁰⁰⁰

Biết 2A=3ⁿ-1. Vậy n=?

Answer 1

a=1+3+3^2+....+3^2000

3a=3(1+3+3^2+....+3^2000)

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(1)

Mà 2a=3^n-1.Từ (1)=>n=2001

Vậy n =2001

Answer 2

Ta có: 3A = 3 + 3^2 + 3^3 + 3^4 +...+3^2001

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

=>  3A - A = 3^2001 - 1

=>  2A = 3^2001 - 1 = 3^n - 1

=> n = 2001

Answer 3

n=2001

Answer 4

n=2001