Question

Cho n là số nguyên dương. Chứng minh rằng:

\(A=2^{3n-1}+2^{3n+1}+1 \) chia hết cho 7

Answer 1

Do n nguyên dương, đặt \(n=m+1\) với m là số tự nhiên

\(\Rightarrow A=2^{3\left(m+1\right)-1}+2^{3\left(m+1\right)+1}+1=2^{3m+2}+2^{3\left(m+1\right)+1}+1\)

\(=4.8^m+2.8^{m+1}+1\)

Do \(8\equiv1\left(mod7\right)\Rightarrow\left\{{}\begin{matrix}8^m\equiv1\left(mod7\right)\\8^{m+1}\equiv1\left(mod7\right)\end{matrix}\right.\)

\(\Rightarrow4.8^m+2.8^{m+1}+1\equiv4+2+1\left(mod7\right)\)

\(\Rightarrow4.8^m+2.8^{m+1}+1⋮7\)