Question

Prove Sn: 2 is a factor of n^2 - n + 2

Answer 1

What I thought would work is do
\[S _{k}:2x =k ^{2}-k+2\]
\[S _{k+1}:2x =(k+1) ^{2}-(k+1)+2\]
Then \[S _{k} = S _{k+1}\]

Answer 2

To me, I would go on this way
if n is odd, then n = 2k +1 for some k in R
then n^2 = (2k+1)^2 = 4k^2+4k +1
therefore \(n^2 -n +2 = 4k^2+4k+1 -2k-1+2=4k^2+2k+2=2(k^2+k+1)\)
hence, n^2 -n +2 divided by2
If n is even, then n = 2m
n^2 = 4m^2
\(n^2 -n+2 = 4m^2 -2m +2 = 2(2m^2-m +1)\)
that shows n^2 -n+2 divided by 2 also.
If n =0, then 2|2
All cases lead to the proof done.

Answer 3

If you want to work it by induction, you need to start by showing that the given statement is true for base case @darielgames

Answer 4

\[S _{k+1}: (k+1) ^{2}-(k+1)+2 = (k^2 - k +2) + 2k\]

Answer 5

From induction hypotheisis \(\large k^2 -k+2 = 2x\)
so

\[S _{k+1}: (k+1) ^{2}-(k+1)+2 = (k^2 - k +2) + 2k = 2x + 2 = 2y\]

ends the proof

Answer 6

I understand what you did right here
(k+1)2−(k+1)+2=(k2−k+2) but why do you add 2k? and where exactly does 2 and 2y come in?

Answer 7

expand \(\large (k+1)^{2}-(k+1)+2 \)

Answer 8

\(\large (k+1)^{2}-(k+1)+2 = k^2 + 2k + 1 - k - 1 + 2\)

yes ?

Answer 9

Alright then the 2 = 2y, you are using y for \[S_{k}\]?

Answer 10

yes thats right
Sorry just making it linear is a bit congusing

Answer 11

confusing

Answer 12

from induction hypothesis (Sk) we know that stuff inside parenthesis is a multiple of 2 right ?

Answer 13

\[\large \begin{align} (k+1)^{2}-(k+1)+2 &= k^2 + 2k + 1 - k - 1 + 2 \\~\\&= (k^2 - k +2) + 2k \\~\\ &= (2x) + 2k\end{align}\]

Answer 14

Sorry I don't understand >.>
What you did was simplification or equality in this one ?
http://puu.sh/bvxIZ/161657d03d.png

Answer 15

Omg.... I get it!

Answer 16

k^2 - k +2 is the base so it's the same thing as saying 2x

Answer 17

@ganeshie8 Thanks!

Answer 18

yw :)
btw, don't forget to prove the statement for base case n=0