1. ## Prove theory

Hi Guys,

Preparing for final and came across home work question.

Question: Prove that n^2 + n is even for any interger n.

Thanks
CHicago boy

2. Hello.

Originally Posted by chicago_boy81
Hi Guys,

Preparing for final and came across home work question.

Question: Prove that n^2 + n is even for any interger n.
Two cases

1) n is odd: then $n = 2m+1, m\in \mathbb{N}_0$

Thus
$n^2 + n = (2m+1)^2 + 2m+1 =(4m^2+2+1)+2m+1 = 4m^2+4+2m = 2(2m^2+4+2m)$

is even

2) n is even: then $n = 2m, m \in \mathbb{N}$

=> $n^2+n = (2m)^2 + 2m = 4m^2 + 2m = 2(2m^2+m)$

Do you understand?

Regards,
Rapha

3. Very easy...

... for $n$ even $n$ and $n^{2}$ are both even, so that $n^{2} + n$ is even...

... for $n$ odd $n$ and $n^{2}$ are both odd, so that $n^{2} + n$ is even...

Kind regards

$\chi$ $\sigma$

4. A third solution won't harm anyone:

$n^2+n = n(n+1)$ and clearly 2 divides one of $n$ or $n+1$.

5. We can conclude remembering the 'legend' according to which Gauss demonstrated when he was seven years old [!] that...

$\sum_{i=1}^{n} i = \frac {n\cdot (n+1)}{2}$ (1)

True or not the 'legend', it is evident from (1) that $n\cdot (n+1)$ must be an even number...

Kind regards

$\chi$ $\sigma$

6. It looks like your proof is going to 2 cases, the 1st case being if n is even so n=2k for some integer k, and the second case will be for if n is odd so n=2m+1 for some integer m. This covers the set all of integers because an integer is either odd or even. hope this helps =).