Questions: Hint: this question is case sensitive, you should ONLY use k, but not K in your answers. Base case n= n^3-4n+6= which is divisible by 3 . Inductive step Assume that for any k 3 , we will prove that 3 . By the inductive hypothesis, there exists an integer m such that =3m. It follows that k^3= . (k+1)^3-4(k+1)+6= k^3+ k^2+ k+ = By inductive hypothesis =3() Since m and k are integers, must be an integer. Hence, 3

Hint: this question is case sensitive, you should ONLY use k, but not K in your answers.
Base case
n= n^3-4n+6= which is divisible by 3 .

Inductive step
Assume that for any k 3 , we will prove that 3 .

By the inductive hypothesis, there exists an integer m such that =3m. It follows that k^3= .
(k+1)^3-4(k+1)+6= k^3+ k^2+ k+
= By inductive hypothesis
=3()

Since m and k are integers, must be an integer. Hence, 3

Transcript text: Hint: this question is case sensitive, you should ONLY use $k$, but not $K$ in your answers. Base case $n=$ $\square$ $n^{3}-4 n+6=$ $\square$ which is divisible by 3 . Inductive step Assume that for any $k$ $\square$ 3| $\square$ , we will prove that $3 \mid$ $\square$ . By the inductive hypothesis, there exists an integer $m$ such that $\square$ $=3 m$. It follows that $k^{3}=$ $\square$ . \[ \begin{aligned} (k+1 & )^{3}-4(k+1)+6=\square k^{3}+\square k^{2}+\square k+\square \\ & =\square \text { By inductive hypothesis } \\ & =3(\square) \end{aligned} \] Since $m$ and $k$ are integers, $\square$ must be an integer. Hence, $3 \mid$ $\square$

Solution

Solution Steps

To solve this problem, we need to use mathematical induction to prove that a given expression is divisible by 3 for all integers $ n $. The process involves two main steps: the base case and the inductive step.

Base Case: Verify that the expression is divisible by 3 for the initial value of $ n $, typically $ n = 0 $ or $ n = 1 $.
Inductive Step: Assume that the expression is divisible by 3 for some integer $ k $. Then, prove that the expression is also divisible by 3 for $ k+1 $.

Step 1: Base Case Verification

To verify the base case, we evaluate the expression $ n^3 - 4n + 6 $ for $ n = 0 $:

\[ 0^3 - 4 \cdot 0 + 6 = 6 \]

Since $ 6 $ is divisible by $ 3 $ (as $ 6 \div 3 = 2 $), the base case holds true.

Step 2: Inductive Hypothesis

Assume that for some integer $ k $, the expression $ k^3 - 4k + 6 $ is divisible by $ 3 $. This means there exists an integer $ m $ such that:

\[ k^3 - 4k + 6 = 3m \]

Step 3: Inductive Step

We need to prove that the expression is also divisible by $ 3 $ for $ k + 1 $. We evaluate the expression for $ n = k + 1 $:

\[ (k + 1)^3 - 4(k + 1) + 6 \]

Expanding this, we have:

\[ (k + 1)^3 = k^3 + 3k^2 + 3k + 1 \] \[ -4(k + 1) = -4k - 4 \] \[ +6 = +6 \]

Combining these, we get:

\[ k^3 + 3k^2 + 3k + 1 - 4k - 4 + 6 = k^3 + 3k^2 - k + 3 \]

Now, we can rearrange this as:

\[ = k^3 - 4k + 6 + 3k^2 - 3 \]

By the inductive hypothesis, we know $ k^3 - 4k + 6 = 3m $. Thus, we can substitute:

\[ = 3m + 3k^2 - 3 = 3(m + k^2 - 1) \]

Since $ m + k^2 - 1 $ is an integer, we conclude that:

\[ (k + 1)^3 - 4(k + 1) + 6 \text{ is divisible by } 3 \]

Final Answer

Since both the base case and the inductive step have been verified, we conclude that the expression $ n^3 - 4n + 6 $ is divisible by $ 3 $ for all integers $ n $.

Thus, the final answer is:

\[ \boxed{\text{The expression is divisible by } 3 \text{ for all integers } n.} \]

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