Question

a=1+3+3^2+3^3+...+3^n

Answer 1

\(3a=3+3^2+3^3+...+3^{n+1}\\ 3a-a=\left(3+3^2+3^3+...+3^{n+1}\right)-\left(1+3+3^2+...+3^n\right)\\ 2a=3^{n+1}-1\\ a=\dfrac{3^{n+1}-1}{2}\)

Answer 2

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

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

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

\(A=\dfrac{3^{n+1}-1}{2}\)

Answer 3

Ta có : A= 1 + 3 + 3² + 3³ + ...+ 3ⁿ

3A = 3 + 3² + 3³ + 3⁴ + ...+ 3ⁿ⁺¹

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

2A = 3ⁿ⁺¹ - 1

A = \(\dfrac{3^{n+1}-1}{2}\)