Question

cho c=1+4+4^1+4^2+4^3+4^4+...+4^20

​chứng minh rằng c chia hết cho 21

Answer 1

\(C=1+4+4^2+4^3+4^4+....+4^{20}\)

\(C=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{18}+4^{19}+4^{20}\right)\)

\(C=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...+4^{18}\left(1+4+4^2\right)\)

\(C=\left(1+4+4^2\right)\left(1+4^3+...+4^{18}\right)\)

\(C=21.\left(1+4^3+...+4^{18}\right)\)

Vì 21 chia hết cho 21 nên \(21.\left(1+4^3+...+4^{18}\right)\) chia hết cho 21(đpcm)

Chúc bạn học tốt!!!

Answer 2

\(C=1+4+4^1+4^2+4^3+4^4+...+4^{20}\)

\(C=\left(1+4+4^2\right)+\left(4^2+4^3+4^4\right)+...+\left(4^{18}+4^{19}+4^{20}\right)\) \(C=\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{18}.\left(1+4+4^2\right)\)

\(C=\left(1+4+4^2\right).\left(1+4^3+...+4^{18}\right)\)

\(C=21.\left(1+4^3+...+4^{18}\right)\)

Vì \(21⋮21\) \(\Rightarrow21.\left(1+4^3+...+4^{18}\right)\)

Vậy \(C⋮21\)