Question

Tính tổng A=4+4²+4³+...+4⁴¹. Chứng minh rằng, A⋮21.

Answer 1

Sửa đề:\(A=4+4^2+4^3+...+4^{21}\)

=>\(4A=4^2+4^3+...+4^{22}\)

=>\(4A-A=4^{22}+4^{21}+...+4^3+4^2-4^{21}-...-4^3-4^2\)

=>\(3A=4^{22}-4^2\)

=>\(A=\dfrac{4^{22}-4^2}{3}\)

\(A=4+4^2+4^3+...+4^{21}\)

\(=\left(4+4^2+4^3\right)+\left(4^4+4^5+4^6\right)+...+\left(4^{19}+4^{20}+4^{21}\right)\)

\(=4\left(1+4+4^2\right)+4^4\left(1+4+4^2\right)+...+4^{19}\left(1+4+4^2\right)\)

\(=21\left(4+4^4+...+4^{19}\right)⋮21\)