Answer:
Let \(\phi(n)=5^{n}(4n+1)-1\).
For \(n=1\), \(\phi(1)=5^1(4\cdot 1+1)-1=5\cdot 5-1=24\), which is divisible by \(8\).
Now assume that for some positive integer \(k\), \(\phi(k)\) is divisible by \(8\). So we may write \(\phi(k)=8r\) for some integer \(r\).
We consider \(\phi(k+1)-\phi(k)\):
\(\phi(k+1)-\phi(k)=5^{k+1}(4k+5)-1-\bigl(5^k(4k+1)-1\bigr)\)
\(=5^{k+1}(4k+5)-5^k(4k+1)\)
\(=5^k\bigl(5(4k+5)-(4k+1)\bigr)=5^k(20k+25-4k-1)\)
\(=5^k(16k+24)=8\cdot 5^k(2k+3)\).
So \(\phi(k+1)-\phi(k)\) is divisible by \(8\). Since \(\phi(k)\) is divisible by \(8\) by the induction hypothesis, it follows that \(\phi(k+1)\) is also divisible by \(8\).
Therefore, by mathematical induction, \(\phi(n)\) is divisible by \(8\) for every positive integer \(n\).
We prove by induction that \(\phi(n)=5^{n}(4n+1)-1\) is divisible by \(8\) for all positive integers \(n\).
Base case. When \(n=1\),
\(\phi(1)=5^1(4\cdot 1+1)-1=5\cdot 5-1=24\).
Since \(24\) is divisible by \(8\), the result is true for \(n=1\).
Inductive step. Assume that for some positive integer \(k\), \(\phi(k)\) is divisible by \(8\). Then there exists an integer \(r\) such that \(\phi(k)=8r\).
Now
\(\phi(k+1)=5^{k+1}(4(k+1)+1)-1=5^{k+1}(4k+5)-1\).
Consider the difference:
\(\phi(k+1)-\phi(k)=\bigl(5^{k+1}(4k+5)-1\bigr)-\bigl(5^k(4k+1)-1\bigr)\).
Simplifying,
\(\phi(k+1)-\phi(k)=5^{k+1}(4k+5)-5^k(4k+1)\)
\(=5^k\bigl(5(4k+5)-(4k+1)\bigr)\)
\(=5^k(20k+25-4k-1)\)
\(=5^k(16k+24)\)
\(=8\cdot 5^k(2k+3)\).
So \(\phi(k+1)-\phi(k)\) is divisible by \(8\).
Since \(\phi(k)\) is divisible by \(8\) by the inductive hypothesis, adding a multiple of \(8\) gives another multiple of \(8\). Hence \(\phi(k+1)\) is divisible by \(8\).
Therefore, by mathematical induction, \(\phi(n)\) is divisible by \(8\) for every positive integer \(n\).
