Question

Cho A =3+3^2+3^3+...+3^100

C/m A  chia hết cho 120

Answer 1

Ta có : A = 3 + 3² + 3³ + 3⁴ + ...... + 3¹⁰⁰

=> A = (3 + 3² + 3³ + 3⁴) + ...... + (3⁹⁷ + 3⁹⁸ + 3⁹⁹ + 3¹⁰⁰)

=> A = (3 + 3² + 3³ + 3⁴) + ...... + 3⁹⁶(3 + 3² + 3³ + 3⁴)

=> A = 120 + ..... + 3⁹⁶.120

=> A = 120(1 + .... + 3⁹⁶) chia hết cho 120

Answer 2

A=\(3+3^2+3^3+3^4+...+3^{100}\)

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

  =\(\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)\)

  =\(\left(3+3^2\right)\left(1+3^2+3^4+...+3^{98}\right)\)

  =\(12\left(1+3^2+3^4+...+3^{98}\right)\)

Vì \(12⋮12\)=>\(12\left(1+3^2+3^4+...+3^{98}\right)⋮12\)

=>\(A⋮12\)

Vậy \(A⋮12\)