Question

CMR: 3^1 + 3^2 +3^3  ...+ 3^99 + 3^100 chia hết cho 4

Answer 1

Ta có: 3¹+3²+3³+…+3⁹⁹+3¹⁰⁰

=(3¹+3²)+(3³+3⁴)+…+(3⁹⁹+3¹⁰⁰)

=3.(1+3)+3³.(1+3)+…+3⁹⁹.(1+3)

=3.4+3³.4+…+3⁹⁹.4

=(3+3³+…+3⁹⁹).4 chia hết cho 4

=>3¹+3²+3³+…+3⁹⁹+3¹⁰⁰ chia hết cho 4

Answer 2

Đặt   \(A=3+3^2+3^3+...+3^{99}+3^{100}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)

\(=3\left(1+3\right)+3^{ 3}\left(1+3\right)+...+3^{99}\left(1+3\right)\)

\(=\left(1+3\right)\left(3+3^3+...+3^{99}\right)\)

\(=4\left(3+3^3+...+3^{99}\right)\)

Vì 4 chia hết cho 4 nên \(4\left(3+3^3+...+3^{99}\right)\)

Vậy A chia hết cho 4