Answer
See explanation
Work Step by Step
Proof by mathematical induction,
Steps:
1. Take any number $k$ and show that the statement holds true for that value of $k$.
2. Show that if the statement is true for $k$, the statement holds true for the value $k+1$. This can be done by taking the statement where the value is $k+1$ and reducing it into the statement where the value is $k$.
Let's start the proof:
Base Case: Show that $P(0)$ is true:
$P(0) = 0^3 – 7\cdot 0 + 3 = 3$ and $3$ is divisible by $3$. So, $P(0)$ is true.
Inductive Case: Show that for all integer $k ≥ 0$, if $P(k)$ is true, then $P(k+1)$ is true:
Let $k$ be any integer such that $k ≥ 0$ and suppose $P(k)$ is true.
This means $k^3 - 7k + 3$ is divisible by $3$. Now we must show that $P(k+1)$ is true.
$P(k+1) = (k+1)^3 - 7(k+1) + 3$
$= k^3 + 3k^2 + 3k +1 -7k -7 + 3$
$= k^3 + 3k^2 -4k -3$
$= k^3 – 7k + 3 +3k^2+3k-6$ [expressing above line in the form of $P(k)$]
$=(k^3-7k+3)+3(k^2+k-2)$
Now we see that $k^3 – 7k + 3$ is equal to $P(k)$ and $P(k)$ is divisible by $3$.
Also $3(k^2+k-2)$ is divisible by $3$.
So $P(k+1)$ is divisible by $3$.
This concludes our proof.
