Question

Cho \(A=3^1+3^2+3^3+...+3^{2006}\)​Thu gọn A Tìm x để 2A+3=3^x

Answer 1

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

Answer 2

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

\(\Rightarrow3A-A=2A=3^{2007}-3^1=3.\left(3^{2006}-1\right)\)

Do đó \(A=\frac{3.\left(3^{2006}-1\right)}{2}\)

Ta có : \(2A+3=3^{2007}-3+3=3^{2007}=3^x\)

\(\Rightarrow x=2007\)

Answer 3

x=

2007

đúng ko

vậy hả

bn

Answer 4

cac  ban giai thich cau a gium minh nhe

Answer 5

trả lời

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

x=2007

Answer 6

+) \(A=3+3^2+3^3+....+3^{2006}\)

\(3A=3\left(3+3^2+3^3+....+3^{2006}\right)\)

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

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

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

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

Vậy \(A=\frac{3^{2007}-3}{2}\)

+) Thay 2A=3²⁰⁰⁷-3 vào 2A+3=3^x ta có:

3²⁰⁰⁷-3+3=3²⁰⁰⁷=3^x

=> x=2007