Question

A= 3+3²+3³+...+3²⁰⁰⁸

Tìm x biết 2A+3=3^x

Answer 1

A = 3 + 3² + 3³ + ... + 3²⁰⁰⁸
3A = 3² + 3³ + 3⁴ + ... + 3²⁰⁰⁹
3A - A = ( 3² + 3³ + 3⁴ + ... + 3²⁰⁰⁹) - ( 3 + 3² + 3³ + ... + 3²⁰⁰⁸)
2A = 3²⁰⁰⁹ - 3
A = \(\frac{3^{2009}-3}{2}\)
\(2A+3=3^x\)
\(\Rightarrow\)\(\frac{3^{2009}-3}{2}\times2+3=3^x\)
\(\Rightarrow3^{2009}-3+3=3^x\)
\(\Rightarrow3^{2009}=3^x\)
\(\Rightarrow x=2009\)

Answer 2

Ta có:3A=3²+3³+.................+3²⁰⁰⁹

\(\Rightarrow\)3A-A=(3²+3³+...............+3²⁰⁰⁹)-(3+3²+3³+................+3²⁰⁰⁸)

\(\Rightarrow2A=3^{2009}-3\)

\(\Rightarrow2A+3=3^{2009}\Rightarrowđpcm\)