Question

Chứng minh :  1 + 3 + 3^2 + 3^3 + .......+3^99 chia hết cho 4

Answer 1

1+3+3²+3³+.....+3⁹⁹

= (1+3)+(3²+3³)+....+(3⁹⁸+3⁹⁹)

= 1(1+3) + 3²(1+3) +.....+ 3⁹⁸(1+3)

= 1.4 + 3².4 +......+ 3⁹⁸.4

= 4.(1+3²+....+3⁹⁸) chia hết cho 4

=> 1+3+3²+3³+.....+3^{99 chia hết cho 4 (đpcm)}

Answer 2

Ta có :

\(1+3+3^2+3^3+...+3^{99}\)

\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{98}+3^{99}\right)\)

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

\(=4+3^2.4+...+3^{98}.4\)

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

Answer 3

Đặt 1+3+3²+3³+3⁴+...+3⁹⁹ = A

A = (1+3)+(3²+3³)+...+(3⁹⁸+3⁹⁹)

A = 4+3².(1+3)+...+3⁹⁸.(1+3)

A = 4+3².4+...+3⁹⁸.4

A = 4.(1+3²+...+3⁹⁸) chia hết cho 4

=> 1+3+3²+...+3⁹⁹ chia hết cho 4