Induction proof verification?

Question

Using induction, I have to prove that $n^2+n$ is divisible by 2. Here's how I did it, and I wanted to know if this is considered valid. I started with $(k+1)^2+(k+1)$, and after simplifying and factoring, I got $(k+2)(k+1)$. From this, can't I just say that if $k$ is odd, then the term $(k+1)$ becomes an even number, and when an even number is multiplied by an odd number, you get an even number which is always divisible by $2$. Then, the same logic can be applied for when $k$ is even and added to the term $(k+2)$? $Q.E.D$...?

Answer 1

There are many ways to prove this statement, but you are instructed to do so by induction. This means you must use the inductive hypothesis, i.e. will require you "capturing" something of the form $k^2+k$.

In this case:

$$(k+1)^2+(k+1) = (k^2+k)+2(k+1)$$

Answer 2

Base Case is trivial.

We assume true for $n=k$, that is: $k^2+k=2m, m\in \Bbb Z$

Then show for $n=k+1$:

$$(k+1)^2+k+1=k^2+2k+1+k+1=k^2+3k+2$$ $$=(k^2+k)+2k+2=2m+2k+2=2(m+k+1)$$

Answer 3

$\begin{align}{\bf Hint}\ \ \ \ f(k) &= (k+1)\,k\\ \Rightarrow\ f(k+1) &= (k+1)(k+2) \end{align}$

$\,\Rightarrow\ f(k)$ & $f(k+1)$ have equal parity since $\,k\,$ & $\,k+2\,$ do. $ $

Therefore $\,f(k)\,$ even $\,\Rightarrow\, f(k+1)\,$ even