>> When n is equal to 1, the left-hand side is 10,
the right-hand side is 2 times 1 times the quantity 6 minus 1,
and that gives us 2 times 5, which is 10.
So the statement is true when n is equal to 1.
Now for your induction hypothesis, which you need to write,
you have to show that you are making an assumption.
So I need the word if or assuming.
Write out the statement everywhere you see an n, write a k.
So that's p sub k, down at the bottom, your p sub k plus 1 statement,
which is 10 plus 6 plus 2 plus, so on, and so forth.
I'm leaving some room.
14 minus 4 times the quantity k plus 1,
and that's equal to 2 times the quantity k plus 1 times the quantity 6 minus the quantity k
plus 1.
What did I miss?
Well, right before the k plus first term would be the kth turn.
Let's write that in there.
So that would be 14 minus 4k, that would be the term right before the k plus first term.
Notice that this series is identical to the series above.
That tells us what we're going to need to add to both sides of our equation.
So now we know we need to add 14 minus 4 times the quantity k plus 1
to both sides of our equation.
Left- hand side matches top and bottom, now we're working on the right-hand side.
Let's look at the bottom first.
We have 2 times the quantity k plus 1 and then we have 6 minus k minus 1.
Let's take that to the side for a second.
We have 6 minus k minus 1.
Well, that gives us 5 minus k. I don't like the negative leading coefficient.
So what we're going to do is we're going to factor that out.
We're going to factor out a negative.
So we would write that as k minus 5.
So we'll have 5 minus k for a minute.
Now of course, we're not going to factor out the negative, we can't put it there.
You can't -- that makes it subtraction.
That's not good.
When we factored out that negative, we're going to have to put that on the negative, on the 2.
So we're going to change that 2 into a negative 2.
And now it's going to be k minus 5.
The bottom's all cleaned.
Now let's work on the top.
We're going to try to match it.
So we're going to go ahead and multiply it out,
because we don't have anything we can factor out.
So we have 12k minus 2k squared plus 14 minus 4k minus 4.
Adding like terms, I have negative 2k squared plus 8k plus 10.
Factor out a negative 2 and we're left with k squared minus 4k minus 5.
All right.
Now let's factor this trinomial.
Plus 1 and minus 5.
There we go.
Therefore, the statement is true for all positive integers n.
Students often ask me, oh, do I need the conclusion?
Can I leave this off?
Can I leave that off?
You can't leave anything off.
You need everything.
Everything I write you need.
Other than maybe the p sub k's, you don't need that stuff.
But everything else is part of the proof, it's part of what you need to provide
for me on the test and on the quizzes.
All right.
For this one, when n is 1, the left-hand side is 1, the right-hand side is 2 to the 1st minus 1.
So that's 2 minus 1, which also gives us 1.
So the statement is true when n is equal to 1.
Our induction hypothesis is 1 plus 2 plus 2 squared plus 2 cubed, and so on and so forth.
Everywhere you see an n, write a k. So we would have 2 to the k minus 1
and that's equal to 2 to the k and then minus 1.
That's our p sub k statement, our induction hypothesis.
Down at the bottom, we're going to write our p sub k plus 1 statement.
Everywhere you seen an n, write a k plus 1.
I left a little room.
Notice here I have 2 to the k plus 1 and then I'm going to subtract 1 from that exponent.
What would come right before 2 to the k plus 1 minus 1?
Well, that would be 2 to the k minus 1.
Well, look, this series is identical to the series up here.
That tells us what we have to add to both sides of this equation in order
to transform the equation up top to the equation down below.
Let's do that.
So we need to add 2 to the k plus 1 minus 1, but what is that?
What is 2 to the k plus 1 minus 1?
Well, that's 2 to the k. There's nothing wrong
with simplifying it before we add it to both sides, so let's do that.
This is simply 2 to the k.
So we're going to add 2 to the k to both sides.
The left-hand sides are identical.
Let's work on the right-hand sides.
So I have one of these and one of these and I have two of these.
So I have 2 times 2 to the k minus 1.
Well, we need to add those exponents, so you have 2 times 2
to the k, so that's an exponent of 1.
So you add those exponents, so that's 2 to the 1 plus k
and that's the same thing as k plus 1, right?
So that's 2 to the k plus 1 minus 1.
We add those exponents.
That does not change the base to a 4.
The base does not change.
You don't change that into a 4.
All right.
Looks good.
Therefore, the statement is true for all positive integers n.