Hi Guys,
Preparing for final and came across home work question.
Question: Prove that n^2 + n is even for any interger n.
Please somebody help urgently.
Thanks
CHicago boy
Hello.
Two cases
1) n is odd: then $\displaystyle n = 2m+1, m\in \mathbb{N}_0$
Thus
$\displaystyle 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 $\displaystyle n = 2m, m \in \mathbb{N} $
=> $\displaystyle n^2+n = (2m)^2 + 2m = 4m^2 + 2m = 2(2m^2+m)$
Do you understand?
Regards,
Rapha
Very easy...
... for $\displaystyle n$ even $\displaystyle n$ and $\displaystyle n^{2}$ are both even, so that $\displaystyle n^{2} + n$ is even...
... for $\displaystyle n$ odd $\displaystyle n$ and $\displaystyle n^{2}$ are both odd, so that $\displaystyle n^{2} + n$ is even...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
We can conclude remembering the 'legend' according to which Gauss demonstrated when he was seven years old [!] that...
$\displaystyle \sum_{i=1}^{n} i = \frac {n\cdot (n+1)}{2}$ (1)
True or not the 'legend', it is evident from (1) that $\displaystyle n\cdot (n+1)$ must be an even number...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
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 =).