Question

A = 2 + 2^2 + 2^3 + 2^4 + ..... + 2^19 + 2^20

chứng tỏ A chia hết cho 3 , chia hết cho 6

Answer 1

\(A=2+2^2+2^3+2^4+...+2^{19}+2^{20}\)

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

\(A=6+6\cdot2^2+...+6\cdot2^{18}\)

\(A=6\cdot\left(1+2^2+...+2^{18}\right)⋮\text{ }3\text{ v}\)

Answer 2

\(A=2+2^2+2^3+....+2^{20}.\)

\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{19}.\left(1+2\right)\)

\(=2.3+2^3.3+...+2^{19}.3\)

\(=3.\left(2+2^3+...+2^{19}\right)⋮3\)

\(\Rightarrow A⋮3\)( đpcm)
\(\text{Vì A có các hạng tử đều là lũy thừa của 2 nên }\) \(A⋮2\)

Vì \(A⋮2\)và \(A⋮3\)Nên \(A⋮6\)(đpcm)

Answer 3

\(A=2+2^2+2^3+....+2^{20}\)

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

\(A=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{19}.\left(1+2\right)\)

\(A=2.3+2^3.3+....+2^{19}.3\)

\(=>...\)