Math Help - Mathematical Inuction problem

1. Mathematical Inuction problem

Having problems with this question

n^3 - 7n + 3 is divisible by 3, for each integer n => 0.

---------------My work-------------------------

Show n is true for n = 0

0^3 -7(0) + 3 = 3

Then show n is true for n = K +1

(K + 1)^3 -7(k + 1) + 3 = k^3 + 3k^2 + 3k +1 - 7K -7 = k^3 + 3k^2 -4k -3

Im kinda stuck on how to proceed...

2. Originally Posted by Thetheorycase
Having problems with this question

n^3 - 7n + 3 is divisible by 3, for each integer n => 0.

---------------My work-------------------------

Show n is true for n = 0

0^3 -7(0) + 3 = 3

Then show n is true for n = K +1

(K + 1)^3 -7(k + 1) + 3 = k^3 + 3k^2 + 3k +1 - 7K -7 = k^3 + 3k^2 -4k -3

Im kinda stuck on how to proceed...
k^3 + 3k^2 -4k=(k-1) k (k+4)

3. Originally Posted by Thetheorycase
Having problems with this question

n^3 - 7n + 3 is divisible by 3, for each integer n => 0.

---------------My work-------------------------

Show n is true for n = 0

0^3 -7(0) + 3 = 3

Then show n is true for n = K +1

(K + 1)^3 -7(k + 1) + 3 = k^3 + 3k^2 + 3k +1 - 7K -7 +3 = k^3 + 3k^2 -4k -3 no, you are oversimplifying!

Im kinda stuck on how to proceed...
You had a small typo (omitted +3).

Continuing...

$k^3+3k^2+3k+1-7k-7+3=\left(k^3-7k+3\right)+3k^2+3k+1-7$

$=\left(k^3-7k+3\right)+3k^2+3k-6$

Now, what can we conclude ?

4. Originally Posted by Archie Meade
You had a small typo (omitted +3).

Continuing...

$k^3+3k^2+3k+1-7k-7+3=\left(k^3-7k+3\right)+3k^2+3k+1-7$

$=\left(k^3-7k+3\right)+3k^2+3k-6$

Now, what can we conclude ?
We conclude that n^3 - 7n + 3 is divisible by 3 meaning: n^3 - 7n + 3 = 3r for some int r

I dont know where i Should sub it?

5. Originally Posted by Thetheorycase
We conclude that n^3 - 7n + 3 is divisible by 3 meaning: n^3 - 7n + 3 = 3r for some int r

I dont know where i Should sub it?
Induction tries to find out if the hypothesis being true for any n=k
causes the hypothesis to be true for n=k+1.

True for n=1 causes true for n=2
True for n=2 causes true for n=3
True for n=3 causes true for n=4 ...... to infinity.

Using k and k+1 examines this cause and effect in general.
If true for k causes true for k+1,
then we only need to examine the first n, as the hypothesis would then be true for all n.

So, if the formula is valid for some n=k,
and being valid for n=k causes validity for n=k+1,
you then only need to test the first term.

This is why you should write the k+1 expression containing the k expression.

If 3 does divide $k^3-7k+3$, then it will definately divide $\left(k^3-7k+3\right)+3\left(k^2+k-2\right)$