Use induction to prove the following statement.

Question

Full Question: Use induction to prove that $3$ divides $2n^3 + n$ for all positive integers $n$.

This is what I have so far. Please correct me if I am wrong and help me finish it thank you.

• For $n = 1$, the assertion says that $3$ divides $2(1)^3 + 1 = 3$, which indeed is the case.

• For the inductive step, I assumed that $3$ divides $2k^3 + k$ for some positive integer $k$. Hence there exists an integer $a$ such $3a = 2k^3 + k$.

• From here I have no clue what to do and where to go.

Thank You.

Answer 1

For the induction step just consider $2(k+1)^3+(k+1)$. We can rewrite the last expresion as $2(k^3+3k^2+3k+1)+(k+1)=(2k^3+k)+3(2k^2+2k+1)$, by the inductive hypothesis $3\mid 2k^3+k$, then $3\mid 2k^3+k+3(2k^2+2k+1)$. So we're done.

Answer 2

Method$\#1:$

If $f(n)=2n^3+n$

$$f(m+1)-f(m)=2(3m^2+3m+1)+m+1-m=3(2m^2+2m+1)$$

So, $3|f(m+1)\iff3|f(m)$

Method$\#2:$

If induction is not mandatory, $$2n^3+n=2\underbrace{(n-1)n(n+1)}_{\text{product of three consecutive integers }}+3n$$