Question

C = 1 + 3 + 3^2 + ........ + 3^11 . CMR C :

a, chia hết cho 13

b, chia hết cho 40

Answer 1

C=(1+3+3²)+(3³+3⁴+3⁵)+...+(3⁹+3¹⁰+3¹¹)

C=13+3³(1+3+3²)+...+3⁹(1+3+3²)

C=13+3³.13+...+3⁹.13

C=13(1+3³+...+3⁹)

Vì nó có thừa số 13 nên chia hết cho 13 (1+3³+...+3⁹ là STN)

C=(1+3+3²+3³)+(3⁴+3⁵+3⁶+3⁷)+(3⁸+3⁹+3¹⁰+3¹¹)

C=40+3⁴(1+3+3²+3³)+3⁸(1+3+3²+3³)

C=40+3⁴.40+3⁸.40

=40(1+3⁴+3⁸)

=>C chia hết cho 40

Answer 2

a) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)

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

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

\(=3\times91+3^7\times91+...+3^{1987}\times91\)

\(=3\times7\times13+3^7\times7\times13+...+3^{1987}\times7\times13\)

\(=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)

Vì \(A=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)nên A chia hết cho 13.

b) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)

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

\(=3\times\left(1+3^2+3^4+3^6\right)+...+3^{1985}\times\left(1+3^2+3^4+3^6\right)\)

\(=3\times820+...+3^{1985}\times820\)

\(=3\times20\times41+...+3^{1985}\times20\times41\)

\(=41\times\left(3\times20+...+3^{1985}\times20\right)\)

Vì \(A=41\times\left(3\times20+...+3^{1985}\times20\right)\)nên A chia hết cho 41.