Question

Cho A= \(3+3^2+3^3+....+3^{2019}\).Tìm số tự nhiên n sao cho \(2A+3=3^n\)

Answer 1

A = 3 + 3² + 3³ +...+3²⁰¹⁹

-> 3A = 3 (3 + 3² + 3³ +...+3²⁰¹⁹)

-> 3A = 3² + 3³ + 3⁴ +...+3²⁰²⁰

-> 3A - A = (3² + 3³ + 3⁴ +...+ 3²⁰²⁰) - (3 + 3² + 3³ +...+3²⁰¹⁹)

-> 2A = 3²⁰²⁰ - 3

\(\rightarrow A=\frac{3^{2020}-3}{2}\)

Ta có: \(2A+3=3^n\)

\(\Rightarrow2\cdot\frac{3^{2020}-3}{2}+3=3^n\)

\(\Rightarrow3^{2020}-3+3=3^n\)

=> 3²⁰²⁰ = 3ⁿ => n = 2020

Answer 2

Trl:

\(A=3+3^2+3^3+...+3^{2018}\)

\(3A=3^2+3^3+3^4+...+3^{2017}+3^{2018}\)

\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{100}+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)

\(\Rightarrow2A=3^{101}-3\)

\(\Rightarrow2A+3=3^{101}\)

\(\Rightarrow n=101\)

Vậy n = 101

Hc tốt

Answer 3

XL , mình làm nhầm đề

Answer 4

A = 3 + 3² + 3³ + ... + 3²⁰¹⁹

3A = 3( 3 + 3² + 3³ + ... + 3²⁰¹⁹ )

      = 3² + 3³ + 3⁴ + ... + 3²⁰²⁰

3A - A = 2A

= ( 3² + 3³ + 3⁴ + ... + 3²⁰²⁰ ) - ( 3 + 3² + 3³ + ... + 3²⁰¹⁹ )

= 3² + 3³ + 3⁴ + ... + 3²⁰²⁰ - 3 - 3² - 3³ - ... - 3²⁰¹⁹

= 3²⁰²⁰ - 3

2A + 3 = 3ⁿ

<=> 3²⁰²⁰ - 3 + 3 = 3ⁿ

=> 3²⁰²⁰ = 3ⁿ

=> n = 2020