Prove that 3ⁿ - 1 is divisible by 2 for all n 1.

For $n=1$, we have $3^1 - 1 = 3 - 1 = 2$.
Since 2 is divisible by 2, $P(1)$ is true.

Assume that $P(k)$ is true for some positive integer $k \ge 1$.
This means $3^k - 1$ is divisible by 2.
So, we can write $3^k - 1 = 2m$ for some integer $m$.
From this, $3^k = 2m + 1$.

We need to prove that $P(k+1)$ is true, i.e., $3^{k+1} - 1$ is divisible by 2.
Consider the expression for $n=k+1$:

Rewrite $3^{k+1}$ as $3 \cdot 3^k$:

Substitute $3^k = 2m + 1$ from the inductive hypothesis:

Distribute the 3:

Factor out 2:

Since $3m+1$ is an integer, $2(3m+1)$ is a multiple of 2.
Therefore, $3^{k+1} - 1$ is divisible by 2.

By the principle of mathematical induction, $P(n)$ is true for all integers $n \ge 1$.
Thus, $3^n - 1$ is divisible by 2 for all $n \ge 1$.

For $n=1$, we have $2^1 = 2$ and $1+1 = 2$.
Since $2 \ge 2$, $P(1)$ is true.

Assume that $P(k)$ is true for some positive integer $k \ge 1$.
This means $2^k \ge k+1$.

We need to prove that $P(k+1)$ is true, i.e., $2^{k+1} \ge (k+1)+1$, which simplifies to $2^{k+1} \ge k+2$.
Consider the left side of the inequality for $n=k+1$:

Rewrite $2^{k+1}$ as $2 \cdot 2^k$:

From the inductive hypothesis, we know $2^k \ge k+1$. Multiply both sides by 2:

We want to show $2^{k+1} \ge k+2$. We have $2^{k+1} \ge 2k+2$.
Now, we need to show that $2k+2 \ge k+2$.
Subtract $k+2$ from both sides:

Since $k \ge 1$, we know $k \ge 0$.
Therefore, $2k+2 \ge k+2$.
Combining the inequalities, we have $2^{k+1} \ge 2k+2 \ge k+2$.
Thus, $2^{k+1} \ge k+2$.

By the principle of mathematical induction, $P(n)$ is true for all integers $n \ge 1$.
Thus, $2^n \ge n+1$ for all $n \ge 1$.

For $n=1$, we have $4^1 - 1 = 4 - 1 = 3$.
Since 3 is divisible by 3, $P(1)$ is true.

Assume that $P(k)$ is true for some positive integer $k \ge 1$.
This means $4^k - 1$ is divisible by 3.
So, we can write $4^k - 1 = 3m$ for some integer $m$.
From this, $4^k = 3m + 1$.

We need to prove that $P(k+1)$ is true, i.e., $4^{k+1} - 1$ is divisible by 3.
Consider the expression for $n=k+1$:

Rewrite $4^{k+1}$ as $4 \cdot 4^k$:

Substitute $4^k = 3m + 1$ from the inductive hypothesis:

Distribute the 4:

Factor out 3:

Since $4m+1$ is an integer, $3(4m+1)$ is a multiple of 3.
Therefore, $4^{k+1} - 1$ is divisible by 3.

By the principle of mathematical induction, $P(n)$ is true for all integers $n \ge 1$.
Thus, $4^n - 1$ is divisible by 3 for all $n \ge 1$.

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^{(