Question

cho A=3+3²+3³+...+3²⁰⁰⁹.tim so tu nhien n biet rang 2A+3=3ⁿ

Answer 1

 

 

\(3A=3^2+3^3+3^4+...+3^{2010}\)

\(3A-A=\left(3^2+3^3+3^4+..+3^{2010}\right)-\left(3+3^2+3^3+....+3^{2009}\right)\)

\(2A=3^{2010}-3\)(1)

 

(1) => \(3^{2010}-3+3=3^{2010}\)

=> n = 2010

 

Answer 2

A = 3 + 3² + 3³ + ... + 3²⁰⁰⁹

3A = 3² + 3³ + 3⁴ + ... + 3²⁰¹⁰

3A - A = (3² + 3³ + 3⁴ + ... + 3²⁰¹⁰) -  (3 + 3² + 3³ + ... + 3²⁰⁰⁹)

2A = 3²⁰¹⁰ - 3

3ⁿ = 2A + 3

3ⁿ = 2²⁰¹⁰ - 3 + 3

3ⁿ = 3²⁰¹⁰

n = 2010