Question

[Nguyễn Qúy Lê Minh]

cho A=3+3³+3⁵+...+3¹⁹⁹¹ .CMR A chia hết cho 13 và 41

help:(

Answer 1

[KAl(SO4)2·12H2O]

Ta có:

\(A=3+3^3+3^5+...+3^{1991}=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+\left(3^{1987}+3^{1989}+3^{1991}\right)\)

\(A=3.\left(1+3^2+3^4\right)+3^7.\left(1+3^2+3^4\right)+...+3^{1987}.\left(3^{1987}+3^{1989}+3^{1991}\right)\)

\(A=3.91+3^7.91+...+3^{1987}.91=3.7.13+3^7.7.13\)

\(A=13.\left(3.7.13+3^7.7+...+3^{1987}.7\right)\)

Vì: \(A=15.\left(2+2^4+...+2^{58}\right)\)nên \(A⋮13\)

Tương tự:

\(A=\left(3+3^3+3^5+3^7\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)

\(A=3.\left(1+3^2+3^4\right)+3^7.\left(1+3^2+3^4\right)+...+3^{1987}.\left(1+3^2+3^4+3^6\right)\)

\(A=3.820+...+3^{1985}.820=3.20.41+...+3^{1985}.20.41\)

\(A=41.\left(3.20+...+3^{1985}.20\right)\)nên \(B⋮41\)

:)

Answer 2

[Tho ngo van]

(3+3^3+3^5)+...+(3^1987+3^1989+3^1991)

=3x(1+3^2+3^4)+...+3^1987x(1+3^2+3^4)

=3x91+...+3^1987x91

=(3+...+3^1987)x91=(3+...+3^1987)x13x7\(⋮\)13

Vậy A\(⋮\)13

(3+3^3+3^5+3^7)+...+(3^1985+3^1987+3^1989+3^1991)

=3x(1+3^2+3^4+3^6)+...+3^1985x(1+3^2+3^4+3^6)

=3x820+...+3^1985x820

=(3+...+3^1985)x820=(3+...+3^1985)x41x20\(⋮\)41

Vậy A\(⋮\)41