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PROFESSOR: Welcome
back to recitation.

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

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do a little bit of practice
with sigma notation.

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So this will be just
a few short problems

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to make sure that you're
comfortable with what

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all the pieces in the
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

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a sum on the left-hand
side, and then

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I've given you two
other sums, but I've

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

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You have enough information
to fill in the two blanks.

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

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is fill in the two blanks
so that the sums are equal.

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And the object is
obviously is to do this

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without writing
out all the terms

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and adding up and
then going backwards.

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So you really want
to try and understand

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what each part of the
sigma notation does.

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The second problem
I'd like you to do

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is a simplification problem.

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There are three finite sums.

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And what I'd like you
to do is combine them

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into a single sum or two sums.

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Do the best you can to get
it as simplified as you can

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without actually
writing out a number,

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but keeping it in some
sort of notation form.

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So the object is just
to combine what you can

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and 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

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what each of these
pieces represents.

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And let's look at the
first sum and make

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sure we 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 second,

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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

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from some number I 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

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we would like on 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 power would be 7.

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

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up here-- based on that
7-- we'd like that power

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to be 2 less 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,

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because 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 here, a 2

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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

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here, the first term
we wanted in this sum,

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was 2 to the first.

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So the first term
we want in this sum

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is going to be 2 to the first.

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

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Another way to
think about this is

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that we know we want the
same number of values

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that we're summing over.

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So notice that from
1 up to 5, there

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are 5 values we're summing over.

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From 3 up to 7,
there are actually

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5 values we're summing over.

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

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but 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,

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it 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

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taken one of the 2's that
was in all of those terms

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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

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to rewrite this in 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 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,

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I need k to be 0 to
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 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,

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the first term is
2 to the first.

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Which is the first term here.

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Let's just check
one more to make

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sure we 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,

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I 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,

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maybe you can expand the
sums and look at them

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and notice that they are
indeed 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've 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

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summing: n cubed minus n
squared, n cubed minus n

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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

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we can split up over
the terms, as long

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as we keep the right index.

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So let me actually use the
regular chalk for this,

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

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to help with the
first and the third.

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So in the second term, notice
I have an n squared and an n

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cubed-- or n cubed
minus n squared here,

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and an n cubed minus
n squared here.

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

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together.

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

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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

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this part of that summation.

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And what's left
in that summation

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is every term I
had 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

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split into 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 have 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

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are being 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

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goes all the way
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

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is 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 terms.

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This has the 45
through 100 terms only.

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So the 45 to 100
terms are in both

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and 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 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 equals 1 to 44

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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 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,

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if I wanted, into a 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,

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I'd have to 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 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,

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this is a sum of five terms.

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I'm going to ask you to
write 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

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and I 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 problem,

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and I see the same kind of
factor in each of these things,

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I like to make it as simple
on myself as possible.

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I like to pull out
that factor just

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to make sure that I
can simplify this as

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much as possible before
I go into 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

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start 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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Now, if you couldn't
do it before,

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you can probably do it now.

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Because now it's
sort of very obvious

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how these terms are changing.

243
00:11:28,534 --> 00:11:30,450
So we want to see how
these terms are changing

244
00:11:30,450 --> 00:11:33,630
and how we could index
them in some variable.

245
00:11:33,630 --> 00:11:37,940
So let's start with the 1/5 and
I'll start with my summation

246
00:11:37,940 --> 00:11:40,250
and then we'll figure out
what all the pieces are.

247
00:11:40,250 --> 00:11:43,390
Now obviously the numerator
in this case is fixed at 1--

248
00:11:43,390 --> 00:11:46,150
and I've got a fraction here,
so the numerator's fixed at 1--

249
00:11:46,150 --> 00:11:48,920
but the sign is alternating.

250
00:11:48,920 --> 00:11:50,380
So how do you alternate sign?

251
00:11:50,380 --> 00:11:54,300
You're going to take negative
1 and raise it to a power.

252
00:11:54,300 --> 00:11:57,310
Now the power you raise
it to will depend on

253
00:11:57,310 --> 00:12:00,010
if you want the first term
to be positive or negative,

254
00:12:00,010 --> 00:12:01,730
and where you start
your summation.

255
00:12:01,730 --> 00:12:03,880
So there's a lot of
choices you can make.

256
00:12:03,880 --> 00:12:09,302
But I'm going to start
my summation, we'll say,

257
00:12:09,302 --> 00:12:12,770
we'll do it in k and
we'll start at k equals 1.

258
00:12:12,770 --> 00:12:15,190
And then we'll have to figure
everything out from that.

259
00:12:15,190 --> 00:12:17,460
So I'm going to start my
summation at k equals 1.

260
00:12:17,460 --> 00:12:20,780
My first term, I want
to be positive 1.

261
00:12:20,780 --> 00:12:25,210
So I need my power
to be k plus 1.

262
00:12:25,210 --> 00:12:28,040
Because now my
power here is going

263
00:12:28,040 --> 00:12:30,575
to be-- when I put in a 1,
I get a 1 plus 1, I get 2.

264
00:12:30,575 --> 00:12:31,950
Negative one
squared is positive.

265
00:12:31,950 --> 00:12:33,230
That's that's what I want.

266
00:12:33,230 --> 00:12:35,940
You might have done k minus 1.

267
00:12:35,940 --> 00:12:37,760
If you did k minus 1, that's OK.

268
00:12:37,760 --> 00:12:40,000
Because k minus 1 is
also an even number.

269
00:12:40,000 --> 00:12:42,430
So when I take negative
1 and I square it,

270
00:12:42,430 --> 00:12:44,350
I still get a positive number.

271
00:12:44,350 --> 00:12:46,580
So there are a lot of choices
one can make and still

272
00:12:46,580 --> 00:12:48,920
be correct on that power.

273
00:12:48,920 --> 00:12:50,730
And then, I'm counting up.

274
00:12:50,730 --> 00:12:53,950
Notice the denominator is
increasing just by 1 each time.

275
00:12:53,950 --> 00:12:58,040
And so it looks like, I could
do just something like over k.

276
00:12:58,040 --> 00:13:00,000
Now let's check if
that makes sense.

277
00:13:00,000 --> 00:13:03,350
Well when k is 1, I get
1 in the denominator.

278
00:13:03,350 --> 00:13:04,480
This is 1 over 1.

279
00:13:04,480 --> 00:13:06,394
When k is 2, I get 2
in the denominator.

280
00:13:06,394 --> 00:13:08,060
When k is 3, I get 3
in the denominator.

281
00:13:08,060 --> 00:13:09,020
So that looks good.

282
00:13:09,020 --> 00:13:12,360
And now the only question is,
where should I stop this thing?

283
00:13:12,360 --> 00:13:14,040
So I have my alternating sign.

284
00:13:14,040 --> 00:13:15,630
My denominator looks right.

285
00:13:15,630 --> 00:13:17,650
For what value of k
do I want to stop?

286
00:13:17,650 --> 00:13:20,740
I want to stop when the
denominator equals 5.

287
00:13:20,740 --> 00:13:23,580
And so I just need
to put a 5 up here.

288
00:13:23,580 --> 00:13:25,550
And then I'm done.

289
00:13:25,550 --> 00:13:28,656
Now, if you wanted to
move the 1/5 back in,

290
00:13:28,656 --> 00:13:29,780
you could actually do that.

291
00:13:29,780 --> 00:13:36,680
Maybe your solution
looked something--

292
00:13:36,680 --> 00:13:38,160
I pull the 1/5 back in.

293
00:13:42,910 --> 00:13:45,450
And I have 5k in there instead.

294
00:13:45,450 --> 00:13:46,854
Maybe that was your solution.

295
00:13:46,854 --> 00:13:48,520
But these are ultimately
the same thing.

296
00:13:48,520 --> 00:13:50,637
Because really this
is just distributing.

297
00:13:50,637 --> 00:13:51,330
Right?

298
00:13:51,330 --> 00:13:52,290
This is a big sum.

299
00:13:52,290 --> 00:13:53,700
I have a 1/5 out in front.

300
00:13:53,700 --> 00:13:56,140
And so I multiply
every term by 1/5.

301
00:13:56,140 --> 00:13:58,342
So I just have to put
a 5 in the denominator.

302
00:13:58,342 --> 00:14:00,300
So you might have had
something more like this.

303
00:14:00,300 --> 00:14:02,340
That's still correct.

304
00:14:02,340 --> 00:14:05,730
So just to stress, that
really the sigma notation,

305
00:14:05,730 --> 00:14:09,572
it's a good tool to understand
how to manipulate easily.

306
00:14:09,572 --> 00:14:11,030
So there are probably
more problems

307
00:14:11,030 --> 00:14:13,377
you can find to practice, if
you're nervous about this.

308
00:14:13,377 --> 00:14:15,960
But I just wanted to give you a
chance to see a couple of them

309
00:14:15,960 --> 00:14:17,630
and how we work on them.

310
00:14:17,630 --> 00:14:18,893
That's where I'll stop.