Question

Tính 

\(A=1+3^2+3^4+...+3^{2006}\)

Answer 1

\(A=1+3^2+3^4+...+3^{2006}\)

\(3^2.A=3^2+3^4+3^6+...+3^{2008}\)

\(3^2.A-A=\left(3^2+3^4+3^6+...+3^{2008}\right)-\left(1+3^2+3^4+...+3^{2006}\right)\)

\(9.A-A=3^{2008}-1\)

\(8.A=3^{2008}-1\)

\(A=\frac{3^{2008}-1}{8}\)

Ủng hộ mk nha ^_-

Answer 2

Ta có A = 1 + 3² + 3⁴ + ...... + 3²⁰⁰⁶

=> 3²A = 3² + 3⁴ + ...... + 3²⁰⁰⁸

=> 9A = 3² + 3⁴ + ...... + 3²⁰⁰⁸

=> 9A - A = (3² + 3⁴ + ...... + 3²⁰⁰⁸) - (1 + 3² + 3⁴ + ...... + 3²⁰⁰⁶)

=> 8A = 3²⁰⁰⁸ - 1

=> A = \(\frac{3^{2008}-1}{8}\)