Show by mathematical induction that 7^(2n+3) + 1 is a multiple of 8 for all positive integers n.

For $n=1$, we have:

Calculate $7^5$:

So, $7^5 + 1 = 16807 + 1 = 16808$.
To check if 16808 is a multiple of 8:

Since 16808 is divisible by 8, $P(1)$ is true.

Assume that $P(k)$ is true for some positive integer $k \ge 1$.
This means $7^{(2k+3)} + 1$ is a multiple of 8.
So, we can write $7^{(2k+3)} + 1 = 8m$ for some integer $m$.
From this, $7^{(2k+3)} = 8m - 1$.

We need to prove that $P(k+1)$ is true, i.e., $7^{(2(k+1)+3)} + 1$ is a multiple of 8.
First, simplify the exponent for $n=k+1$:

So we need to show $7^{(2k+5)} + 1$ is a multiple of 8.
Consider the expression for $n=k+1$:

Rewrite $7^{(2k+5)}$ as $7^2 \cdot 7^{(2k+3)}$:

Substitute $7^{(2k+3)} = 8m - 1$ from the inductive hypothesis:

Distribute the 49:

Factor out 8:

Since $49m - 6$ is an integer, $8(49m - 6)$ is a multiple of 8.
Therefore, $7^{(2(k+1)+3)} + 1$ is a multiple of 8.

By the principle of mathematical induction, $P(n)$ is true for all positive integers $n$.
Thus, $7^{(2n+3)} + 1$ is a multiple of 8 for all positive integers $n$.

For $n=1$, we check if $P(1)$ is true.
Given $u_1 = 1$.
Using the proposed formula $u_n = 3^n - 2$:

Since the value from the formula matches the given initial condition ($u_1 = 1$), $P(1)$ is true.

Assume that $P(k)$ is true for some positive integer $k \ge 1$.
This means we assume $u_k = 3^k - 2$.

We need to prove that $P(k+1)$ is true, i.e., $u_{k+1} = 3^{k+1} - 2$.
From the given recurrence relation, we know:

Substitute the inductive hypothesis $u_k = 3^k - 2$ into this equation:

Distribute the 3:

This is exactly the formula for $u_{k+1}$ that we wanted to prove.
Thus, $P(k+1)$ is true.

By the principle of mathematical induction, the statement $u_n = 3^n - 2$ is true for all positive integers $n \ge 1$.