\(n=2k+1\)
\(\Rightarrow A=1+2.4^k+3.9^k+4.16^k+5.25^k\)
- Ta có: \(4\equiv1\left(mod3\right)\Rightarrow2.4^k\equiv2mod\left(3\right)\)
\(16\equiv1\left(mod3\right)\Rightarrow4.16^k\equiv1\left(mod3\right)\)
\(25\equiv1\left(mod3\right)\Rightarrow5.25^k\equiv2\left(mod3\right)\)
\(\Rightarrow A\equiv\left(1+2+1+2\right)\left(mod3\right)\Rightarrow A⋮3\)
Tương tự ta có:
\(A\equiv\left(1+-2-3+4\right)\left(mod5\right)\Rightarrow A⋮5\)
Mà 3 và 5 nguyên tố cùng nhau \(\Rightarrow A⋮15\)