PROFESSOR: Welcome
back to recitation.
In this video, what
I'd like us to do is,
do a little bit of practice
with sigma notation.
So this will be just
a few short problems
to make sure that you're
comfortable with what
all the pieces in the
sigma notation actually do.
We're going to start
with two problems here.
And the first one is going to
be a fill-in-the-blanks type
of problem.
And the object
is, I've given you
a sum on the left-hand
side, and then
I've given you two
other sums, but I've
left in each place two blanks,
and I've filled in the rest.
You have enough information
to fill in the two blanks.
So what I'd like you
to do in this problem
is fill in the two blanks
so that the sums are equal.
And the object is
obviously is to do this
without writing
out all the terms
and adding up and
then going backwards.
So you really want
to try and understand
what each part of the
sigma notation does.
The second problem
I'd like you to do
is a simplification problem.
There are three finite sums.
And what I'd like you
to do is combine them
into a single sum or two sums.
Do the best you can to get
it as simplified as you can
without actually
writing out a number,
but keeping it in some
sort of notation form.
So the object is just
to combine what you can
and simplify where you can.
And then we'll do another
one in a little bit.
But first let's do these two.
I'll give you a while to work
on them and then I'll be back.
OK, welcome back.
We're going to start
with the first problem.
So the idea is to
really understand
what each of these
pieces represents.
And let's look at the
first sum and make
sure we understand
what's going on.
So we have 2 raised to a power.
And what we do is we
index over k from 1 to 5.
So we're going to take 2 to the
first, plus 2 to the second,
all the way up to
2 to the fifth.
And that's where the sum stops.
Now in this summation,
k is indexed
from some number I haven't
told you yet, up to 7.
And I didn't specify
what power of k we want.
So there are a couple ways
you can think about this.
It's maybe easiest to work
from what we have up here.
We know that the
exponent, last exponent
we would like on the power
of 2 is a 5 in the end.
But right now, if we just put
a k here, the power would be 7.
So what we'd like to do is
make whatever the power is
up here-- based on that
7-- we'd like that power
to be 2 less than the number
that we're putting in there.
So probably we would like
this to be a k minus 2,
because notice then, the last
number you put in, you get a 5.
Which corresponds to the last
number you put in here, a 2
to the fifth.
Now the last number
here is 2 to the fifth.
And this now will dictate what
we put in the blank down here.
Because the first
value of k we wanted
here, the first term
we wanted in this sum,
was 2 to the first.
So the first term
we want in this sum
is going to be 2 to the first.
So that means that we
would like k to start at 3.
Another way to
think about this is
that we know we want the
same number of values
that we're summing over.
So notice that from
1 up to 5, there
are 5 values we're summing over.
From 3 up to 7,
there are actually
5 values we're summing over.
You might think there are
4, because 7 minus 3 is 4,
but you actually have
to count: 3, 4, 5, 6, 7.
You see in fact there
are 5 values there.
So don't get confused by that.
The differences are the same.
5 minus 1 is 4.
7 minus 3 is 4.
So that's good.
We have the same number of
things we're summing up over.
And the first
terms are the same.
And then you notice, because
of the way we've written,
it actually is going
to be exactly equal.
You could expand and check.
but these are going
to be equal sums.
Now the third one, I was
a little trickier, maybe.
I pulled out a factor of 2.
And so now what
we've done is we've
taken one of the 2's that
was in all of those terms
and we pulled it out.
Right?
So what do we have here?
Well we still have 2 to the k.
But what does this
actually equal?
To make it easier
on myself, I'm going
to rewrite this in another way.
If I pull the 2 back in,
I get a 2 to the k plus 1.
So now what I've done is I've
given you this 2 pulled out.
What it's actually doing is it's
changing the exponent value.
But again, what do we want
the exponents to run over?
We want them to start,
this exponent to start at 1
and to end at 5.
So to get it to start
at 1 and end at 5,
I need k to be 0 to
start, and finish at 4.
And that will be sufficient.
Now again, let's just make sure
that this makes sense to us.
If k is 0, I get
2 to the 0 here.
But when I multiply
by a 2 in front,
the first term is
2 to the first.
Which is the first term here.
Let's just check
one more to make
sure we feel good about it.
When k equals 1, I get a 2
to the first here, times 2.
So that's a 2 squared.
That's the second term
in this sum is 2 squared.
The second term in this sum
is when I put in k equals 2,
I get a 2 squared.
So we see in fact that I've
chosen these values in blue.
Now these three sums
are actually equal.
If you're still
nervous about it,
maybe you can expand the
sums and look at them
and notice that they are
indeed going to work.
Now what I'd like us to do is
work on simplifying a problem.
And if you'll notice, I've put
in three sums, the values here,
two of them are from 1 to 100,
one of them is from 45 to 100.
And the three different
things that I'm
summing: n cubed minus n
squared, n cubed minus n
squared minus n, and then n.
And I wanted us to simplify
this as much as we could.
Now because these
are finite sums
we can split up over
the terms, as long
as we keep the right index.
So let me actually use the
regular chalk for this,
and I'm going to look at how
I can split up the second term
to help with the
first and the third.
So in the second term, notice
I have an n squared and an n
cubed-- or n cubed
minus n squared here,
and an n cubed minus
n squared here.
So what I can do is, I'm
going to look at those terms
together.
And then I'm going to look at
the n, the terms-- or summation
with n and the summation with n.
And we'll compare them.
So let me write out what we get.
We're going to leave the first
one alone for the moment.
And then I'm going
to subtract off
this part of that summation.
And what's left
in that summation
is every term I
had a minus n also.
So I'm going to pull that
minus out with this negative.
And what I'm doing is I'm
taking 45-- n equals 45 to 100
of these added up.
And then n equals 45 to
100 of this added up.
So I end up with another term.
n equals 45 to 100 of just n.
So those two terms are
coming from the middle one
split into two pieces.
And then the last term,
I just write down.
So now, it's set up
to go nicely for this
into a single summation.
And this into a
single summation.
And then we'll see if we
can combine them further.
So if you look here, I have
n equal 1 to a 100 of a sum.
And then I have n equal 45
to a 100 of the same sum.
What's that actually mean?
That means I'm seeing the 45
to 100 thing here, and here.
And there's a difference.
Right?
So I plug in n equals 45 here.
I get 45 to the third
minus 45 squared.
I plug in n equals 45
here, I get the same thing.
And I'm subtracting.
So what's actually happening
is all the terms that have,
that show up in both
this sum and this sum
are being subtracted off.
What are those terms?
Those are all the terms
for n equal 45 up to 100.
Because it's in this
summation and this one
goes all the way
from 1 to a 100.
So it certainly
includes 45 to 100.
So, in fact, you see that all
you wind up with in the end
is n equals 1 to 44 of
n cubed minus n squared.
Again why is that?
This has the 1 through 44 terms
and it has the 45 to 100 terms.
This has the 45
through 100 terms only.
So the 45 to 100
terms are in both
and they're subtracted off.
So that's one way to think
about why we end up with n
equals 1 to 44 of this sum.
And then let's look
what we get here.
Well in fact, we see it's
exactly the same kind of thing.
This is 45 to 100.
This is 1 to 100.
But notice now that the minus
is on the 1 to 100 part.
So I'm actually going to get
negative of n equals 1 to 44
of n.
Because the 45 terms here,
45 to 100, are in both.
So the 45 to 100 here, subtract
off the 45 to 100 here.
Those all go away.
But I'm still left with
the minus n equals 1 to 44.
And now I could
simplify this further,
if I wanted, into a single sum.
1 to 44 n cubed minus
n squared minus n.
Why can I do that so easily?
They're indexing
over the same values.
That's an important point.
If this was indexing
over different values,
I'd have to change this formula
in order to substitute it in.
But because they're indexing
over exactly the same values,
I can just take these two pieces
and put them into a single sum.
So we're going to stop
those two problems now.
We're going to do one more
summation notation problem.
So we're going to
come over here.
And I'm just going
to ask you to write,
this is a sum of five terms.
I'm going to ask you to
write this in sigma notation.
And the main thing-- there will
be multiple ways to do this.
So you might come up with a
different answer than I do.
But I'd like you to work
on it for a few minutes.
And then when you feel
confident, come back
and I will show you how
I solved the problem.
OK, welcome back one more time.
We're going to try and put
this in sigma notation.
And I have to tell you that when
I look at this kind of problem,
and I see the same kind of
factor in each of these things,
I like to make it as simple
on myself as possible.
I like to pull out
that factor just
to make sure that I
can simplify this as
much as possible before
I go into sigma notation.
So the common factor
to all of these is 1/5.
I'm going to pull
out a 1/5 before I
start doing anything else.
There I get a 1.
There I get a minus 1/2
plus 1/3 minus 1/4 plus 1/5.
Now, if you couldn't
do it before,
you can probably do it now.
Because now it's
sort of very obvious
how these terms are changing.
So we want to see how
these terms are changing
and how we could index
them in some variable.
So let's start with the 1/5 and
I'll start with my summation
and then we'll figure out
what all the pieces are.
Now obviously the numerator
in this case is fixed at 1--
and I've got a fraction here,
so the numerator's fixed at 1--
but the sign is alternating.
So how do you alternate sign?
You're going to take negative
1 and raise it to a power.
Now the power you raise
it to will depend on
if you want the first term
to be positive or negative,
and where you start
your summation.
So there's a lot of
choices you can make.
But I'm going to start
my summation, we'll say,
we'll do it in k and
we'll start at k equals 1.
And then we'll have to figure
everything out from that.
So I'm going to start my
summation at k equals 1.
My first term, I want
to be positive 1.
So I need my power
to be k plus 1.
Because now my
power here is going
to be-- when I put in a 1,
I get a 1 plus 1, I get 2.
Negative one
squared is positive.
That's that's what I want.
You might have done k minus 1.
If you did k minus 1, that's OK.
Because k minus 1 is
also an even number.
So when I take negative
1 and I square it,
I still get a positive number.
So there are a lot of choices
one can make and still
be correct on that power.
And then, I'm counting up.
Notice the denominator is
increasing just by 1 each time.
And so it looks like, I could
do just something like over k.
Now let's check if
that makes sense.
Well when k is 1, I get
1 in the denominator.
This is 1 over 1.
When k is 2, I get 2
in the denominator.
When k is 3, I get 3
in the denominator.
So that looks good.
And now the only question is,
where should I stop this thing?
So I have my alternating sign.
My denominator looks right.
For what value of k
do I want to stop?
I want to stop when the
denominator equals 5.
And so I just need
to put a 5 up here.
And then I'm done.
Now, if you wanted to
move the 1/5 back in,
you could actually do that.
Maybe your solution
looked something--
I pull the 1/5 back in.
And I have 5k in there instead.
Maybe that was your solution.
But these are ultimately
the same thing.
Because really this
is just distributing.
Right?
This is a big sum.
I have a 1/5 out in front.
And so I multiply
every term by 1/5.
So I just have to put
a 5 in the denominator.
So you might have had
something more like this.
That's still correct.
So just to stress, that
really the sigma notation,
it's a good tool to understand
how to manipulate easily.
So there are probably
more problems
you can find to practice, if
you're nervous about this.
But I just wanted to give you a
chance to see a couple of them
and how we work on them.
That's where I'll stop.
