Question

Chứng minh rằng 

A = 3 + 3² + 3³ + .... + 3⁹⁹ + 3¹⁰⁰ chia hết cho 4

Answer 1

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

\(A=3.1+3.3+3^3.1+3^3.3+....+3^{98}.1+3^{98}.3\)

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

\(A=3.4+3^3.4+....+3^{98}.4\)

\(A=4.\left(3+3^3+....+3^{98}\right)\) Chia hết cho 4 

Answer 2

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Answer 3

A = 3 + 3² + 3³ + .... + 3⁹⁹ + 3¹⁰⁰

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

A= 3.( 1 + 3 ) + 3³.( 1 + 3 )+...+ 3⁹⁹.( 1 + 3 )

A=4.(3 + 3³ +...+ 3⁹⁹) chia hết cho 4(vì 4 chia hết cho 4)

hay A = 3 + 3² + 3³ + .... + 3⁹⁹ + 3¹⁰⁰ chia hết cho 4

Answer 4

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