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PROFESSOR: Welcome back
to recitation.

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In this video, what I'd like us
to do is, do a little bit

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of practice with
sigma notation.

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So this will be just a few short
problems to make sure

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that you're comfortable with
what all the pieces in the

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sigma notation actually do.

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We're going to start with
two problems here.

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And the first one is going to
be a fill-in-the-blanks type

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of problem.

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And the object is, I've given
you a sum on the left-hand

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side, and then I've given you
two other sums, but I've left

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in each place two blanks, and
I've filled in the rest. You

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have enough information to
fill in the two blanks.

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So what I'd like you to do in
this problem is fill in the

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two blanks so that the
sums are equal.

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And the object is obviously is
to do this without writing out

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all the terms and adding up
and then going backwards.

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So you really want to try and
understand what each part of

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the sigma notation does.

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The second problem I'd
like you to do is a

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simplification problem.

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There are three finite sums. And
what I'd like you to do is

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combine them into a single sum
or two sums. Do the best you

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can to get it as simplified as
you can without actually

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writing out a number, but
keeping it in some sort of

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notation form.

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So the object is just to combine
what you can and

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simplify where you can.

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And then we'll do another
one in a little bit.

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But first let's do these two.

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I'll give you a while to work on
them and then I'll be back.

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OK, welcome back.

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We're going to start with
the first problem.

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So the idea is to really
understand what each of these

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pieces represents.

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And let's look at the first
sum and make sure we

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understand what's going on.

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So we have 2 raised
to a power.

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And what we do, is we index
over k from 1 to 5.

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So we're going to take 2 to
the first, plus 2 to the

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second, all the way up
to 2 to the fifth.

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And that's where
the sum stops.

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Now in this summation, k is
indexed from some number I

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haven't told you yet, up to 7.

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And I didn't specify what
power of k we want.

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So there are a couple ways
you can think about this.

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It's maybe easiest to work from
what we have up here.

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We know that the exponent, last
exponent we would like on

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the power of 2 is
a 5 in the end.

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But right now, if we just
put a k here, the

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power would be 7.

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So what we'd like to do is make
whatever the power is up

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here-- based on that 7-- we'd
like that power to be 2 less

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than the number that we're
putting in there.

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So probably we would like this
to be a k minus 2, because

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notice then, the last number
you put in, you get a 5.

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Which corresponds to the
last number you put in

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here, a 2 to the fifth.

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Now the last number here
is 2 to the fifth.

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And this now will dictate what
we put in the blank down here.

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Because the first value of k we
wanted here, the first term

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we wanted in this sum, was 2
to the first. So the first

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term we want in this sum is
going to be 2 to the first. So

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that means that we would
like k to start at 3.

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Another way to think about this
is that we know we want

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the same number of values
that we're summing over.

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So notice that from 1 up to 5,
there are 5 values we're

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summing over.

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From 3 up to 7, there are
actually 5 values we're

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summing over.

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You might think there are 4,
because 7 minus 3 is 4, but

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you actually have to count:
3, 4, 5, 6, 7.

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You see in fact there
are 5 values there.

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So don't get confused by that.

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The differences are the same.

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5 minus 1 is 4.

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7 minus 3 is 4.

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So that's good.

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We have the same number of
things we're summing up over.

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And the first terms
are the same.

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And then you notice, because of
the way we've written, it

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actually is going to
be exactly equal.

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You could expand and check.

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but these are going
to be equal sums.

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Now the third one, I was a
little trickier, maybe.

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I pulled out a factor of 2.

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And so now what we've done is
we've taken one of the 2's

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that was in all of those terms
and we pulled it out.

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Right?

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So what do we have here?

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Well we still have 2 to the k.

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But what does this
actually equal?

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To make it easier on myself, I'm
going to rewrite this in

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another way.

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If I pull the 2 back in, I
get a 2 to the k plus 1.

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So now what I've done is I've
given you this 2 pulled out.

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What it's actually doing
is it's changing

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the exponent value.

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But again, what do we want the
exponents to run over?

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We want them to start, this
exponent to start at 1

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and to end at 5.

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So to get it to start at 1 and
end at 5, I need k to be 0 to

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start, and finish at 4.

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And that will be sufficient.

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Now again, let's just
make sure that this

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makes sense to us.

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If k is 0, I get 2
to the 0 here.

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But when I multiply by a 2 in
front, the first term is 2 to

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the first. Which is the
first term here.

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Let's just check one more
to make sure we

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feel good about it.

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When k equals 1, I get a 2 to
the first here, times 2.

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So that's a 2 squared.

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That's the second term in
this sum is 2 squared.

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The second term in this sum is
when I put in k equals 2, I

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get a 2 squared.

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So we see in fact that I've
chosen these values in blue.

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Now these three sums
are actually equal.

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If you're still nervous about
it, maybe you can expand the

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sums and look at them and notice
that they are indeed

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going to work.

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Now what I'd like us to do is
work on simplifying a problem.

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And if you'll notice, I put in
three sums, the values here,

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two of them are from 1 to 100,
one of them is from 45 to 100.

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And the three different things
that I'm summing: n cubed

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minus n squared, n cubed minus n
squared minus n, and then n.

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And I wanted us to simplify
this as much as we could.

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Now because these are finite
sums we can split up over the

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terms, as long as we keep
the right index.

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So let me actually use the
regular chalk for this, and

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I'm going to look at how I can
split up the second term to

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help with the first
and the third.

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So in the second term,
notice I have an n

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squared and an n cubed--

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or n cubed minus n squared here,
and an n cubed minus n

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squared here.

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So what I can do is,
I'm going to look

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at those terms together.

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And then I'm going to look at
the the n, the terms, or the

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summation with n and the
summation with n.

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And we'll compare them.

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So let me write out
what we get.

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We're going to leave the first
one alone for the moment.

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And then I'm going to
subtract off this

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part of that summation.

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And what's left in that
summation is every term I had

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a minus n also.

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So I'm going to pull that minus
out with this negative.

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And what I'm doing is I'm taking
45, n equals 45 to 100

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of these added up.

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And then n equals 45 to
100 of this added up.

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So I end up with another term.

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n equals 45 to 100 of just n.

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So those two terms are coming
from the middle one split into

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two pieces.

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And then the last term,
I just write down.

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So now, it's set up to
go nicely for this

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into a single summation.

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And this into a single
summation.

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And then we'll see if we can
combine them further.

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So if you look here, I have n
equal 1 to a 100 of a sum.

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And then I n equal 45 to
a 100 of the same sum.

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What's that actually mean?

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That means I'm seeing the 45 to
100 thing here, and here.

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And there's a difference.

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Right?

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So I plug in n equals 45 here.

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I get 45 to the third
minus 45 squared.

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I plug in n equals 45 here,
I get the same thing.

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And I'm subtracting.

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So what's actually happening
is all the terms that have,

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that show up in both this sum
and this sum are being

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subtracted off.

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What are those terms?

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Those are all the terms for
n equal 45 up to 100.

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Because it's in this summation
and this one goes all the way

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from 1 to a 100.

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So it certainly includes
45 to 100.

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So, in fact, you see that all
you wind up with in the end is

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n equals 1 to 44 of n cubed
minus n squared.

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Again why is that?

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This has the 1 through 44 terms
and it has the 45 to 100

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terms. This has the 45 through
100 terms only.

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So the 45 to 100 terms
are in both and

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they're subtracted off.

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So that's one way to think about
why we end up with n

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equals 1 to 44 of this sum.

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And then let's look
what we get here.

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Well in fact, we see
it's exactly the

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same kind of thing.

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This is 45 to 100.

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This is 1 to 100.

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But notice now that the minus
is on the 1 to 100 part.

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So I'm actually going
to get negative of n

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equals 1 to 44 of n.

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Because the 45 terms here,
45 to 100, are in both.

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So the 45 to 100 here,
subtract off

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the 45 to 100 here.

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Those all go away.

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But I'm still left with the
minus n equals 1 to 44.

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And now I could simplify this
further, if I wanted, into a

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single sum.

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1 to 44 n cubed minus
n squared minus n.

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Why can I do that so easily?

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They're indexing over
the same values.

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That's an important point.

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If this was indexing over
different values, I'd have to

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change this formula in order
to substitute it in.

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But because they're indexing
over exactly the same values,

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I can just take these two pieces
and put them into a

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single sum.

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So we're going to stop those
two problems now.

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We're going to do one more
summation notation problem.

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So we're going to
come over here.

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And I'm just going to ask you
to write, this is a sum of

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five terms. I'm going
to ask you to write

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this in sigma notation.

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And the main thing, there will
be multiple ways to do this.

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So you might come up with a
different answer than I do.

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But I'd like you to work on
it for a few minutes.

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And then when you feel
confident, come back and I

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will show you how I solved
the problem.

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OK, welcome back
one more time.

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We're going to try and put
this in sigma notation.

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And I have to tell you that when
I look at this kind of

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problem, and I see the same
kind of factor in each of

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these things, I like to
make it as simple

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on myself as possible.

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I like to pull out that factor
just to make sure that I can

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simplify this as much as
possible before I go into

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sigma notation.

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So the common factor to
all of these is 1/5.

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I'm going to pull out
a 1/5 before I start

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doing anything else.

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There I get a 1.

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There I get a minus 1/2 plus
1/3 minus 1/4 plus 1/5.

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00:11:18,140 --> 00:11:21,100

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00:11:21,100 --> 00:11:22,740
Now, if you couldn't do
it before, you can

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00:11:22,740 --> 00:11:23,710
probably do it now.

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00:11:23,710 --> 00:11:27,010
Because now it's sort
of very obvious how

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00:11:27,010 --> 00:11:28,750
these terms are changing.

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00:11:28,750 --> 00:11:30,800
So we want to see how these
terms are changing and how we

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00:11:30,800 --> 00:11:33,630
could index them in
some variable.

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00:11:33,630 --> 00:11:37,380
So let's start with the 1/5
and I'll start with my

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00:11:37,380 --> 00:11:38,810
summation and then we'll
figure out what

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00:11:38,810 --> 00:11:40,250
all the pieces are.

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00:11:40,250 --> 00:11:43,390
Now obviously the numerator in
this case is fixed at 1--

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00:11:43,390 --> 00:11:44,640
and I've got a fraction
here, so the

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00:11:44,640 --> 00:11:46,150
numerator's fixed at 1--

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00:11:46,150 --> 00:11:48,920
but the sign is alternating.

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00:11:48,920 --> 00:11:50,380
So how do you alternate sign?

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00:11:50,380 --> 00:11:54,300
You're going to take negative
1 and raise it to a power.

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00:11:54,300 --> 00:11:57,850
Now the power you raise it to
will depend on if you want the

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00:11:57,850 --> 00:12:01,080
first term to be positive or
negative, and where you start

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00:12:01,080 --> 00:12:01,730
your summation.

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00:12:01,730 --> 00:12:03,880
So there's a lot of choices
you can make.

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00:12:03,880 --> 00:12:09,735
But I'm going to start my
summation, we'll say, we'll do

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00:12:09,735 --> 00:12:12,770
it in k and we'll start
at k equals 1.

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00:12:12,770 --> 00:12:15,190
And then we'll have to figure
everything out from that.

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00:12:15,190 --> 00:12:17,460
So I'm going to start my
summation at k equals 1.

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00:12:17,460 --> 00:12:20,780
My first term, I want
to be positive 1.

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00:12:20,780 --> 00:12:25,210
So I need my power
to be k plus 1.

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00:12:25,210 --> 00:12:29,030
Because now my power here is
going to be-- when I put in a

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00:12:29,030 --> 00:12:30,720
1, I get a 1 plus 1, I get 2.

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00:12:30,720 --> 00:12:31,950
Negative one squared
is positive.

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00:12:31,950 --> 00:12:33,230
That's that's what I want.

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00:12:33,230 --> 00:12:35,940
You might have done k minus 1.

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00:12:35,940 --> 00:12:37,760
If you did k minus
1, that's OK.

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00:12:37,760 --> 00:12:40,000
Because k minus 1 is also
an even number.

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00:12:40,000 --> 00:12:42,970
So when I take negative 1 and
I square it, I still get a

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00:12:42,970 --> 00:12:44,350
positive number.

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00:12:44,350 --> 00:12:46,700
So there are a lot of choices
one can make and still be

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00:12:46,700 --> 00:12:48,920
correct on that power.

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00:12:48,920 --> 00:12:50,730
And then, I'm counting up.

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00:12:50,730 --> 00:12:52,380
Notice the denominator
is increasing

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00:12:52,380 --> 00:12:53,950
just by 1 each time.

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00:12:53,950 --> 00:12:58,040
And so it looks like, I could do
just something like over k.

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00:12:58,040 --> 00:13:00,000
Now let's check if
that makes sense.

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00:13:00,000 --> 00:13:03,350
Well when k is 1, I get
1 in the denominator.

287
00:13:03,350 --> 00:13:04,480
This is 1 over 1.

288
00:13:04,480 --> 00:13:06,440
When k is 2, I get 2
in the denominator.

289
00:13:06,440 --> 00:13:08,060
When k is 3, I get 3
in the denominator.

290
00:13:08,060 --> 00:13:09,020
So that looks good.

291
00:13:09,020 --> 00:13:10,310
And now the only question
is, where

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00:13:10,310 --> 00:13:12,360
should I stop this thing?

293
00:13:12,360 --> 00:13:14,040
So I have my alternating sign.

294
00:13:14,040 --> 00:13:15,630
My denominator looks right.

295
00:13:15,630 --> 00:13:17,650
For what value of k
do I want to stop?

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00:13:17,650 --> 00:13:20,740
I want to stop when the
denominator equals 5.

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00:13:20,740 --> 00:13:23,580
And so I just need to
put a 5 up here.

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00:13:23,580 --> 00:13:25,550
And then I'm done.

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00:13:25,550 --> 00:13:29,000
Now, if you wanted to move the
1/5 back in, you could

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00:13:29,000 --> 00:13:29,780
actually do that.

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00:13:29,780 --> 00:13:36,680
Maybe your solution
looked something--

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00:13:36,680 --> 00:13:38,160
I pull the 1/5 back in.

303
00:13:38,160 --> 00:13:42,910

304
00:13:42,910 --> 00:13:45,450
And I have 5k in
there instead.

305
00:13:45,450 --> 00:13:46,910
Maybe that was your solution.

306
00:13:46,910 --> 00:13:48,520
But these are ultimately
the same thing.

307
00:13:48,520 --> 00:13:50,637
Because really this is
just distributing.

308
00:13:50,637 --> 00:13:51,330
Right?

309
00:13:51,330 --> 00:13:52,290
This is a big sum.

310
00:13:52,290 --> 00:13:53,700
I have a 1/5 out in front.

311
00:13:53,700 --> 00:13:56,140
And so I multiply every
term by 1/5.

312
00:13:56,140 --> 00:13:58,780
So I just have to put a
5 in the denominator.

313
00:13:58,780 --> 00:14:00,300
So you might have had something
more like this.

314
00:14:00,300 --> 00:14:02,340
That's still correct.

315
00:14:02,340 --> 00:14:05,990
So just to stress, that really
the sigma notation, it's a

316
00:14:05,990 --> 00:14:09,580
good tool to understand how
to manipulate easily.

317
00:14:09,580 --> 00:14:11,500
So there are probably more
problems you can find to

318
00:14:11,500 --> 00:14:13,760
practice, if you're nervous
about this.

319
00:14:13,760 --> 00:14:15,680
But I just wanted to give you
a chance to see a couple of

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00:14:15,680 --> 00:14:17,630
them and how we work on them.

321
00:14:17,630 --> 00:14:19,260
That's where I'll stop.

322
00:14:19,260 --> 00:14:19,393