Question

Prove that n^2+5n+2 is an even integer for all n<-N

Answer 1

what does n<-N mean?

Answer 2

if you meant for n in the natural numbers
then you can do this by induction

Answer 3

if n=1 then n^2+5n+2=? <--is this even
Now assume for some integer k>=1 k^2+5k+2 can be expressed as 2i
we want to show (k+1)^2+5(k+1)+2 can also be expressed as 2l

l and i are integers by the way

Answer 4

you can also do this without induction

Answer 5

if you know 2|n(n+1)

Answer 6

\[n^2+5n+2 \\ n^2+n+4n+2 \\ n(n+1)+4n+2\]
so if you know 2|n(n+1)
then you can play with that way and it isn't so bad :)

Answer 7

this is what i meant
\[n^2+5n+2, n \ \leftarrow \mathbb{N}\]

Answer 8

If i use induction, do i do two cases, even and odd or jut even?
@freckles

Answer 9

we are trying to show it is even
so we are trying to show n^2+5n+2 is even for any integer n
and so we only need to show it is just for n=1
then assume it is true for n=k
and show it is true for n=k+1
so yes it is a one case type of proof

Answer 10

\[(k+1)^2+5(k+1)+2\]
hint here do a little multiplying
so you use k^2+5k+2=2i

Answer 11

okay i got it. thanks

Answer 12

np