Question 1075272: Use mathematical induction to prove that the statement is true for every positive integer n. Show your work.
2 is a factor of n2 - n + 2
pls give step by step answer
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Use mathematical induction to prove that the statement is true for every positive integer n. Show your work.
2 is a factor of n2 - n + 2
pls give step by step answer
-----
1st:: Show it is true for n = 1::
1^2 - 1 + 2 = 2
true
-------------------
2nd:: Assume it is true for n = k::
2 is a factor of k^2-k+2
-------
3rd:: Then prove is must be true for n = k+1
(k+1)^2 - (k+1) +2 = k^2 + 2k+1 -k-1+2
= [k^2 -k +2] + [2k+1-1]
= [k^2-k+2]+[2k]
Divisible be 2 because k^2-k+2 is divisible be 2
and 2k is divisible by 2.
====
Cheers,
Stan H.
----------------
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website!
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
|
|
|
