Question

1+3+3²+3³+...+3²⁰⁰⁰

Answer 1

 

Đặt BT trên là A ta có

\(3A=3+3^2+3^3+3^4+...+3^{2000}+3^{2001}\)

\(2A=3A-A=3^{2001}-1\)

\(A=\frac{3^{2001}-1}{2}\)

Answer 2

A= 1+3+3²+3³+...+3²⁰⁰⁰

3A= 3+3²+3³+3⁴+.. .+3²⁰⁰¹

3A-A=(3+3²+3³+3⁴+.. .+3²⁰⁰¹)-(1+3+3²+3³+...+3²⁰⁰⁰)

2A= 3²⁰⁰¹-1

A =(3²⁰⁰¹-1) :2