Use mathematical induction to prove the statement is true for all positive integers n. The integer n^3 + 2n is divisible by 3 for every positive integer n.

Question

Use mathematical induction to prove the statement is true for all positive integers n.  The integer n^3 + 2n is divisible by 3 for every positive integer n.

anonymous · Answer

just for clarification...thats $n^3+2n$  :)

anonymous · Answer

@mukushla yeah sorry!

anonymous · Answer

Well, consider $n^3+2n=n(n^2+2)$. If $3\,|\,n$ it's obvious. If $3\not|\,n$ we have $n\equiv1\text{ or }n\equiv 2\pmod3$ hence $n^2\equiv1\pmod3$ and thus $n^2+2\equiv0\mod3$... so we're done :-) sorry this isn't an inductive proof but I wanted to play with modular arithmetic.

Anyways, where were we? Let's assume our statement is true for some positive integer $k$, i.e. $k^3+2k$ is divisible by $3$. Anyways, consider then $k+1$ and we have:$$(k+1)^3+2(k+1)=k^3+3k^2+3k+1+2k+2=(k^3+2k)+3(k^2+k+1)$$; clearly since $3$ divides $k^3+2k,3(k^2+k+1)$ it follows that $3$ divides $(k+1)^3+2(k+1)$. Now, we merely show it starts for the first positive integer $n=1$ and we've proved it all for all integers:$$1^3+2(1)=1+2=3$$and $3$ divides $3$. It follows that $n^3+2n$ is divisible by $3$ for all positive integers $n$

anonymous · Answer

Thanks! I wasn't sure where to go after k^3+3k^2+5k+3 c:

anonymous · Answer

@Adri1580 generally since you're assuming $k^3+2k$ is divisible by $3$ you want to those terms 'out' and consider the remaining ones to show it holds for $(k+1)^3+2(k+1)$ as well :-)

anonymous · Answer

Thanks again for everything :)

anonymous · Answer

move* those terms 'out' and to the side so you can focus on the new remaining ones**