Question

cho a = 3+3²+3³+...+3⁶⁰

a) a chia hết cho 4                                        b) a chia hết cho 13

Answer 1

b) \(A=3+3^2+3^3+...+3^{60}\)

\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)

\(A=3\left(1+3+3^2\right)+3^4\cdot\left(1+3+3^2\right)+...+3^{58}\cdot\left(1+3+3^2\right)\)

\(A=3\cdot13+3^4\cdot13+...+3^{58}\cdot13\)

\(A=13\cdot\left(3+3^4+...+3^{58}\right)\)

Vậy A chia hết cho 13

Answer 2

a) \(A=3+3^2+...+3^{60}\)

\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)

\(A=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+...+3^{59}\cdot\left(1+3\right)\)

\(A=4\cdot\left(3+3^3+...+3^{59}\right)\)

Nên A chia hết cho 4