Question

cho C = 1 + 3 +\(3^2+3^3+....+3^{11}\)

chứng minh rằng a, C chia hết cho 13                   b, C  chia hết cho   40

Answer 1

Ta có : \(3C=3+3^2+3^3+......+3^{12}\)
\(\Rightarrow3C-C=\left(3+3^2+3^3+....+3^{12}\right)-\left(1+3+3^2+3^3+...+3^{11}\right)=3^{12}-1=531440\)
 \(hoặc\)\(2C=531140\Rightarrow C=265720\)chia hết cho 13 và 40

Answer 2

b, \(C=1+3+3^2+3^3+...+3^{11}\)

      \(=\left(1+3+3^2+3^3\right)+...+\left(3^8+3^9+3^{10}+3^{11}\right)\)

       \(=\left(1+3+9+27\right)+...+3^8.\left(1+3+3^2+3^3\right)\)

        \(=40+...+3^8.40\)

         \(=40.\left(1+...+3^8\right)⋮40\)

\(\Rightarrow\) \(C⋮40\)