Question

[Hoàng Quỳnh Mai]

cho A=3+3^2+3^3+.................................+3^9+3^10 .chứng minh A chia hết cho 4

giúp mik nhanh nhé mn

Answer 1

[Phạm Nguyễn Tất Đạt]

\(A=3+3^2+3^3+3^4+...+3^9+3^{10}\)(có 10 số)

\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^9+3^{10}\right)\)(có 5 nhóm)

\(A=3\left(1+3\right)+3^3\left(1+3\right)+...+3^9\left(1+3\right)\)

\(A=\left(1+3\right)\left(3+3^3+...+3^9\right)\)

\(A=4\left(3+3^3+...+3^9\right)⋮4\left(đpcm\right)\)

Answer 2

[Quốc Đạt]

A = 3+3²+3³+...+3⁹+3¹⁰

A = (3+ 3²)+(3³+3⁴)+...+(3⁹+3¹⁰)

A = 3(1+3)+3³(1+3)+...+3⁹ (1+3)

A = (1+3)(3+3³+...+3⁹)

A = 4(3+3³+...+3⁹) => chia hết cho 4