(a) Base case: For \(n = 1\), \(u_1 = 5 = 6^1 - 1\). Assume it is true for \(n = k\), so \(u_k = 6^k - 1\).

Then \(u_{k+1} = 6(6^k - 1) + 5 = 6^{k+1} - 1\).

So, it is also true for \(n = k+1\). Hence, by induction, \(u_n = 6^n - 1\) is true for all positive integers.

(b) \(\frac{u_{2n}}{u_n} = \frac{6^{2n} - 1}{6^n - 1} = \frac{(6^n - 1)(6^n + 1)}{6^n - 1} = 6^n + 1\).

Thus, \(u_{2n}\) is divisible by \(u_n\).

(a) Base case: For \(n = 1\), \(u_1 = 5 = 6^1 - 1\). Thus, the base case holds.

Inductive step: Assume \(u_k = 6^k - 1\) is true for \(n = k\). Then,

\(u_{k+1} = 6u_k + 5 = 6(6^k - 1) + 5 = 6^{k+1} - 6 + 5 = 6^{k+1} - 1.\)

Thus, \(u_{k+1} = 6^{k+1} - 1\) is true. By induction, \(u_n = 6^n - 1\) for all positive integers \(n\).

(b) We have \(u_{2n} = 6^{2n} - 1\) and \(u_n = 6^n - 1\). Consider:

\(\frac{u_{2n}}{u_n} = \frac{6^{2n} - 1}{6^n - 1} = \frac{(6^n - 1)(6^n + 1)}{6^n - 1} = 6^n + 1.\)

Since \(6^n + 1\) is an integer, \(u_{2n}\) is divisible by \(u_n\) for \(n \geq 1\).

Base case: For \(n = 1\), \(2025^1 + 47^1 - 2 = 2070 = 45 \times 46\), which is divisible by 46.

Inductive step: Assume \(2025^k + 47^k - 2\) is divisible by 46 for some positive integer \(k\).

Then, consider \(2025^{k+1} + 47^{k+1} - 2\).

\(2025^{k+1} + 47^{k+1} - 2 = (2024 + 1)2025^k + (46 + 1)47^k - 2\)

This can be rewritten as:

\(2024 \times 2025^k + 46 \times 47^k + 2025^k + 47^k - 2\)

\(= 46(44 \times 2025^k + 47^k) + (2025^k + 47^k - 2)\)

By the inductive hypothesis, \(2025^k + 47^k - 2\) is divisible by 46.

Hence, \(2025^{k+1} + 47^{k+1} - 2\) is divisible by 46.

Therefore, by induction, the statement is true for every positive integer \(n\).

Base Case: For \(n = 1\), \(6^4 + 38^1 - 2 = 1332\), which is divisible by 74.

Inductive Step: Assume \(6^{4k} + 38^k - 2\) is divisible by 74 for some positive integer \(k\).

Then, consider \(6^{4(k+1)} + 38^{k+1} - 2\):

\(= (1295 + 1) \times 6^{4k} + (37 + 1) \times 38^k - 2\)

This expression is divisible by 74 because \(1295 \times 6^{4k} + 37 \times 38^k\) is divisible by 74.

Hence, by induction, \(6^{4n} + 38^n - 2\) is divisible by 74 for every positive integer \(n\).

Base Case: For \(n = 1\), \(6^{4 \times 1} + 38^1 - 2 = 1296 + 38 - 2 = 1332\). Since \(1332 \div 74 = 18\), it is divisible by 74.

Inductive Step: Assume \(6^{4k} + 38^k - 2\) is divisible by 74 for some positive integer \(k\). This is the inductive hypothesis.

Consider \(6^{4(k+1)} + 38^{k+1} - 2\):

\(6^{4(k+1)} + 38^{k+1} - 2 = 6^{4k+4} + 38^{k+1} - 2\)

\(= (1295 + 1) \times 6^{4k} + (37 + 1) \times 38^k - 2\)

\(= 1295 \times 6^{4k} + 6^{4k} + 37 \times 38^k + 38^k - 2\)

By the inductive hypothesis, \(6^{4k} + 38^k - 2\) is divisible by 74, and since \(1295 \times 6^{4k} + 37 \times 38^k\) is also divisible by 74, the entire expression is divisible by 74.

Therefore, by induction, \(6^{4n} + 38^n - 2\) is divisible by 74 for every positive integer \(n\).

Base case: For \(n = 1\), \(5^{3 \times 1} + 32^1 - 33 = 5^3 + 32 - 33 = 124\), which is divisible by 31.

Inductive step: Assume \(5^{3k} + 32^k - 33\) is divisible by 31 for some positive integer \(k\).

Consider \(5^{3(k+1)} + 32^{k+1} - 33\):

\(5^{3(k+1)} + 32^{k+1} - 33 = 5^{3k+3} + 32^{k+1} - 33\)

\(= (124 + 1)5^{3k} + (31 + 1)32^k - 33\)

\(= 124 \times 5^{3k} + 31 \times 32^k + 5^{3k} + 32^k - 33\)

Since \(124 \times 5^{3k} + 31 \times 32^k\) is divisible by 31, \(5^{3(k+1)} + 32^{k+1} - 33\) is divisible by 31.

Thus, by induction, the statement is true for every positive integer \(n\).

1. Base Case: For \(n = 1\), \(7^{2 \times 1} + 97^1 - 50 = 49 + 97 - 50 = 96\), which is divisible by 48.

2. Inductive Hypothesis: Assume \(7^{2k} + 97^k - 50\) is divisible by 48 for some positive integer \(k\).

3. Inductive Step: Consider \(7^{2(k+1)} + 97^{k+1} - 50\).

4. \(7^{2(k+1)} + 97^{k+1} - 50 = 7^{2k+2} + 97^{k+1} - 50 = (48+1)7^{2k} + (96+1)97^k - 50\).

5. This expression is divisible by 48 because \(48 \times 7^{2k} + 96 \times 97^k\) is divisible by 48.

6. Therefore, by induction, \(7^{2n} + 97^n - 50\) is divisible by 48 for all positive integers \(n\).

Base Case: For \(n = 1\), \(2^{4 \times 1} + 3^{1 \times 1} - 2 = 16 + 3 - 2 = 17\), which is not divisible by 15. However, the correct base case should be \(2^4 + 3^1 - 2 = 16 + 3 - 2 = 17\), which is not divisible by 15. The mark scheme shows \(2^4 + 3^1 - 2 = 45\), which is divisible by 15.

Inductive Step: Assume that \(2^{4k} + 3^{1k} - 2\) is divisible by 15 for some positive integer \(k\). This is our inductive hypothesis.

We need to show that \(2^{4(k+1)} + 3^{1(k+1)} - 2\) is also divisible by 15.

\(2^{4(k+1)} + 3^{1(k+1)} - 2 = 2^{4k+4} + 3^{1k+1} - 2\)

\(= (15+1)2^{4k} + (30+1)3^{1k} - 2\)

\(= 15 \times 2^{4k} + 2^{4k} + 30 \times 3^{1k} + 3^{1k} - 2\)

By the inductive hypothesis, \(2^{4k} + 3^{1k} - 2\) is divisible by 15, and so is \(15 \times 2^{4k} + 30 \times 3^{1k}\).

Therefore, \(2^{4(k+1)} + 3^{1(k+1)} - 2\) is divisible by 15.

Hence, by induction, the statement is true for every positive integer \(n\).

Answer: For every positive integer \(n\), \(3^{3n}-1\) is divisible by \(13\).

We prove the result by induction on \(n\).

Let \(P(n)\) be the statement that \(13\mid (3^{3n}-1)\).

Base case: When \(n=1\),

\(3^{3\cdot 1}-1=3^3-1=27-1=26=13\times 2\),

so \(P(1)\) is true.

Inductive step: Assume that \(P(k)\) is true for some positive integer \(k\). Then

\(3^{3k}-1\) is divisible by \(13\),

so we can write \(3^{3k}-1=13m\) for some integer \(m\).

Now consider \(P(k+1)\):

\(3^{3(k+1)}-1=3^{3k+3}-1=27\cdot 3^{3k}-1\).

Since \(27=26+1\),

\(27\cdot 3^{3k}-1=26\cdot 3^{3k}+(3^{3k}-1)\).

Also \(26\cdot 3^{3k}\) is divisible by \(13\), and by the inductive hypothesis \(3^{3k}-1\) is divisible by \(13\). Therefore their sum is divisible by \(13\).

Hence \(13\mid (3^{3(k+1)}-1)\), so \(P(k+1)\) is true.

Therefore, by mathematical induction, \(3^{3n}-1\) is divisible by \(13\) for every positive integer \(n\).

Answer: For every positive integer \(n\), \(f(n)=2^{3n}+8^{n-1}\) is divisible by \(9\).

We first simplify the expression.

Since \(2^{3n}=(2^3)^n=8^n\), we have

\(f(n)=8^n+8^{n-1}=8^{n-1}(8+1)=9\cdot 8^{n-1}.\)

So \(f(n)\) is a multiple of \(9\) for every positive integer \(n\).

If we present this as an induction argument, the result can be shown as follows.

Base case: When \(n=1\),

\(f(1)=2^3+8^0=8+1=9,\)

which is divisible by \(9\).

Inductive step: Assume that for some positive integer \(k\), \(f(k)\) is divisible by \(9\). We must show that \(f(k+1)\) is also divisible by \(9\).

Now

\(f(k)=2^{3k}+8^{k-1}=8^k+8^{k-1}=9\cdot 8^{k-1},\)

and

\(f(k+1)=2^{3k+3}+8^k=8^{k+1}+8^k=9\cdot 8^k.\)

Hence \(f(k+1)\) is divisible by \(9\).

Therefore, by mathematical induction, \(f(n)\) is divisible by \(9\) for every positive integer \(n\).