Question

Given that 'n' is a natural number. Prove the following text with mathematical induction:
2n^3-3n^2+n+6 is a multiple of 6.
Please show your work too. Thanks!

Answer 1

Let

Given statement be P(n)

$\\ \sf\longmapsto P(n)=2n^3-3n^2+n+6,is\:a\:multiple\:of\:6$

For n=1

$\\ \sf\longmapsto P(1)=2(1)^3-3(1)^2+1+6=2-3+7=-1+7=6$

• For n=1 P(n) is true

Now

P(k) will be true

$\\ \sf\longmapsto P(k)=2k^3-3k^2+k+6=6m$

• m means multiple .

We shall prove it for P(k+1)

$\\ \sf\longmapsto P(k+1)$

$\\ \sf\longmapsto 2(k+1)^3-3(k+1)^2+(k+1)+6$

$\\ \sf\longmapsto 2(k^3+3k^2+3k+1)-3(k^2+2k+1)+k+1+6$

$\\ \sf\longmapsto 2k^3+6k^2+6k+2-3k^2-6k-3+k+7$

$\\ \sf\longmapsto 2k^3+6k^2-3k^2+6k-6k+k+2-3+7$

$\\ \sf\longmapsto 2k^3+3k^2+k-1+7$

$\\ \sf\longmapsto 2k^3+3k^2+k+6$

$\\ \sf\longmapsto 2k^3-3k^2+k+6(-1)$

$\\ \sf\longmapsto -mq$

Here

• q stands for any multiple of 6

Hence.

$\\ \sf\longmapsto P(k+1)=-mq=$

• -mq is also a multiple of 6

Thus

• P(k+1) is also true .

Hence from the principle of mathematical induction P(n) is true for n$\epsilon$N.

Answer 2

6 is of course a multiple of 6, so you need only focus on $2n^3-3n^2+n$.

When n = 1, this expression has a value of 2 - 3 + 1 = 0, which is indeed a multiple of 6 (and any natural number for that matter).

Assume that 6 divides $2k^3-3k^2+k$.

Now,

$2(k+1)^3 - 3(k+1)^2 + (k+1) = (2k^3 + 6k^2 + 6k + 2) - (3k^2 + 6k + 3) + (k + 1) \\\\ = 2k^3 + 3k^2 + k \\\\ = 2k^3 - 3k^2 + k + 6k$

and this is divisible by 6. (6k is obviously a multiple of 6; that 6 divides the other three terms is due to the induction hypothesis.)