Question

cho A = 4 + 4 mũ 2 + 4 mũ 3 + 4 mũ  4 + ....... + 4 mũ 99 + 4 mũ 100

chứng tỏ rằng A chia hết cho 5

Answer 1

Ta có:

A = 4 + 4²  + 4³ + 4⁴ + ... + 4⁹⁹ + 4¹⁰⁰

A = (4 + 4²) + (4³ + 4⁴) + ... + (4⁹⁹ + 4¹⁰⁰)

A = 4(1 + 4) + 4³(1 + 4) + ... + 4⁹⁹(1 + 4)

A = 4.5 + 4³.5 + ... + 4⁹⁹.5

A = 5.(4 + 4³ + ... + 4⁹⁹)

Vậy A chia hết cho 5

Answer 2

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

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

\(A=4.\left(1+4\right)+4^3.\left(1+4\right)+...+4^{99}.\left(1+4\right)\)

\(A=4.5+4^3.5+..4^{99}.5\)

\(A=5.\left(4+4^3+...4^{99}\right)\)

\(\Rightarrow A⋮5\)

Answer 3

A=4+4²+4³+4⁴+......+4⁹⁹+4¹⁰⁰

=> A=(4+4²)+(4³+4⁴)+......+(4⁹⁹+4¹⁰⁰)

=> A=4(1+4)+4³(1+4)+.....+4⁹⁹(1+4)

=> A=4.5+4³.5+.....+4⁹⁹.5

=> A=5(4+4³+....+4⁹⁹)

=> A chia hết cho 5 (đpcm)

Answer 4

Ta có:

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

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

\(A=4.\left(1+4\right)+4^3.\left(1+4\right)+...+4^{99}.\left(1+4\right)\)

\(A=4.5+4^3.5+...+4^{99}.5\)

\(A=5.\left(4+4^3+...+4^{99}\right)\)

Vì \(5⋮5\Rightarrow(4+4^3+...+4^{99})⋮5\)

\(\Rightarrow A⋮5\)

Vậy \(A⋮5\)