Question

cho A = 3 + 3² + 3³ +  ... + 3²⁰⁰⁶

a) thu gọn A

b) tìm x để 2A + 3 = 3^x

Answer 1

3A - A = (3² + 3³ + 3⁴ + ... + 3²⁰⁰⁷) - (3 + 3² + 3³ + ... + 3²⁰⁰⁶)

2A = 3²⁰⁰⁷ - 3\(\Rightarrow\hept{\begin{cases}A=\frac{3^{2007}-3}{2}\\2A+3=3^{2007}\Rightarrow x=2007\end{cases}}\)

Answer 2

\(A=3+3^2+3^3+...+3^{2016}\)

\(\Rightarrow3A=3\left(3+3^2+3^3+...+3^{2016}\right)\)

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2017}\)

\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+3^{2016}\right)\)

\(\Rightarrow2A=-3+3^{2017}\)

\(\Rightarrow A=\frac{3+3^{2017}}{2}\)

b) \(2A+3=-3+3-3^{2017}=3^{2017}=3^x\)

\(\Rightarrow x=2017\)

Answer 3

mình cũng ko biết câu này

Answer 4

\(3A=3^2+3^3+3^4+...+3^{2006}+3^{2007}\)

\(3A-A=3^{2007}-3\)

\(A=\frac{3^{2007}-3}{2}\)

ta có:

2\(2.\frac{3^{2007}-3}{2}+3+3^x\)

\(3^{2007}=3^x\)

=>\(x=2007\)