Question

cho B = 4¹ + 4² + 4³  +  ..... + 4³⁰⁰ chứng minh rằng B chia hết cho 5

Answer 1

\(B=4^1+4^2+4^3+...+4^{300}\)

\(B=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{299}+4^{300}\right)\)

\(B=4\left(1+4\right)+4^3\left(1+4\right)+...+4^{299}\left(1+4\right)\)

\(B=4.5+4^3.5+...+4^{299}.5\)

\(B=5\left(4+4^3+...+4^{299}\right)\)

Có : \(B=5\left(4+4^3+...+4^{299}\right)⋮5\)

\(\Rightarrow B⋮5\)

Answer 2

Ta có B= (4¹+4²)+(4³+4⁴)+.....+(4²⁹⁹+4³⁰⁰) 

          B= 4¹(1+4)+4³(1+4)+...+4²⁹⁹(1+4)

           B= 5.(4¹+4³+...+4²⁹⁹)

vì 5 chia hết cho 5 => B chia hết cho 5

Answer 3

A = 1+4+4^2+4^3+...+4^98
= (1+4+4^2) + (4^3+4^4+4^5) + ... + (4^96+4^97+4^98)
=  (1+4+16) + 4^3.(1+4+4^2)+ ... + 4^96.(1+4+4^2)
= 21 + 4^3.21 + ... + 4^96.21
= 21.(1+4^3+..+4^96) chia hết cho 21