Question

Use mathematical induction to prove that the statement is true for every positive integer n.

2 is a factor of n^2 - n + 2

Answer 1

n^2-2n+n+2
n*(n-2)+1*(n-2)
(n-2)*(n+1)
n=2,1
so 2 is the factor

Answer 2

thank you @speedymath

Answer 3

First check if the statement is true for n = 0.

\[\bf (0)^2-(0)+2=2\]So that works.

Now assume that the statement is true for when n = k. Now we must prove the case for n = k + 1.

\[\bf (k+1)^2-(k+1)+2=k^2+k+2\]Now it should be noted that if k is even, i.e. k = 2n, then the expression is always true. So now we must prove that the statement is true if k was odd, i.e. k = 2n + 1\[\bf (2n+1)^2+(2n+1)+2=4n^2+6n+4=2(2n^2+3n+2)\]Clearly, we can see that when k is odd, the statement still holds as 2 is a factor of the final expression.

Hence 2 is a factor of n^2 - n + 2.

@ska8terchick_

Answer 4

wow thank you @genius12 your name do you justice

Answer 5

@ska8terchick_
fan/medal? lol -.-