Hi!
So Im not allowed to use induction on this one.
How do i prove that n^2 + n is even when n is natural number?
Not a clue on this one. Would you mind to help a bit?
A number can be represent in two forms:
n=2k with $\displaystyle k\in\mathbb{Z}$
or
n=2k+1 with $\displaystyle k\in\mathbb{Z}$
First case we have:
$\displaystyle (2k)^2+2k=4k^2+2k=2(2k^2+k)$
so, $\displaystyle 2|(n^2+n)$ for all even n.
Second case:
$\displaystyle (2k+1)^2+(2k+1)=4k^2+4k+2=2(2k^2+2k+1)$
so, $\displaystyle 2|(n^2+n)$ for all odd n.
thus, $\displaystyle 2|(n^2+n) \ \forall n\in\mathbb{Z}$
Hello : There is the solution
I'have two cases:
1 case : $\displaystyle n \equiv 0\left[ {\bmod 2} \right]$
$\displaystyle \left\{ \begin{array}{l}
n \equiv 0\left[ {\bmod 2} \right] \\
n^2 \equiv 0\left[ {\bmod 2} \right] \\
\end{array} \right. \Rightarrow n^2 + n \equiv 0\left[ {\bmod 2} \right]$
2 case: $\displaystyle n \equiv 1\left[ {\bmod 2} \right]
$
$\displaystyle \left\{ \begin{array}{l}
n \equiv 1\left[ {\bmod 2} \right] \\
n^2 \equiv 1\left[ {\bmod 2} \right] \\
\end{array} \right. \Rightarrow n^2 + n \equiv 0\left[ {\bmod 2} \right]
$
By Contradiction:
If $\displaystyle n$ is even there is nothing to prove, so assume $\displaystyle n$ is odd and $\displaystyle n^2+n$ is also odd.
As $\displaystyle n$ is odd it can be written $\displaystyle n=2k+1$ for some natural $\displaystyle k$, then:
$\displaystyle n^2+n=4k^2+4k+1+2k+1= ... $
Hence ... a contradiction.
CB
Hello : you have any contradiction , because I’can written :
$\displaystyle \begin{array}{l}
n^2 + n = 4k^2 + 4k + 1 + 2k + 1 \\
n^2 + n = 2\left( {2k^2 + 2k + 1 + k} \right) \\
n^2 + n = 2\alpha \\
\alpha = 2k^2 + 2k + 1 + k \\
\end{array}
$
Conclusion : even number