Question 1075272
<pre>
First let's see what Pk+1 would be: 

[That's always the first thing to do.  Before you start an
induction proof,  you should calculate Pk+1 to see where 
you're headed]: 

To do that, replace n by k+1 in n²-n+2 to see what the Pk+1 
is, for that is what we are going for, and if we have that 
beforehand, we'll know when we have arrived and the proof 
is finished. 

Substituting k+1 for n in n²-n+2,
we have 

Pk+1:  2 is a factor of (k+1)²-(k+1)+2 = k²+2k+1-k-1+2 = k²+k+2.  

Now that we know what Pk+1 is, we know where we're going, and 
we'll know we have arrived if and when we get that 2 is a factor 
of k²+k+2.  

Now we can start the proof: 

P1:  substitute n=1 into n²-n+2 and get 1²-1+2 = 2, and 2 is 
indeed a factor of 2.  So P1 true. 

Assume Pk:  that is, 2 is a factor of k²-k+2.  

Now if we add an even number to an even number we get an even 
number.  So we add the even number 2k to it and we get k²-k+2+2k 
or k²+k+2.  So 2 being a factor of k²-k+2 implies that 2 is a 
factor of k²+k+2.  

This is, Pk implies Pk+1 and P1 is true, so our induction proof is 
now complete.

Edwin</pre>