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Let s sub k for k is equal
to 1, to all the way to 100, denote the sum of the infinite
geometric series whose first term is k minus
1 over k factorial. And the common
ratio is 1 over k. Then the value of 100 squared
over 100 factorial plus the sum from k equals 1 to 100
of the absolute value of this thing over here, k
squared minus 3k plus 1 times s sub k is. So this looks like a
pretty daunting thing. But I guess a good
place to start would just to be find a
simple way to express s sub k. So if we were to
write s sub k is equal to the first term is
k minus 1 over k factorial. And then the common
ratio is 1 over k. So it's going to be 1 over k
to the 0 plus 1 over k to the 1 plus 1 over k squared. It's just going to
keep going on forever. It is an infinite
geometric series. Now there is a very
well-known formula for the sum of an
infinite geometric series. I enjoy the proof of
it so much that we're going to do it again. And I've done it
multiple times already, but it's such a simple
proof that it's always good to start from
basic principles. If you're doing this
under time pressure, you could just use the formula. But if we say, let's
just say some other sum. Let's call this thing just s. s is equal to 1 over k to
the 0 plus 1 over k to the 1 plus 1 over k squared all
the way, keep going on into infinity. Now let's multiply
1 over k times s. 1 over k times s is just going
to be 1 over k times each of these terms. When I multiply 1 over
k times this term, I'm going to get 1 over
k to the first power. When I multiply 1 over
k times this term, I'm going to get 1 over
k to the second power all the way on to infinity. So if I were to
subtract this from that. So if I were to subtract that--
so this coefficient is 1. So it's 1. So we have 1 minus
1 over k times s-- I'm just
subtracting that-- would be equal to all of this stuff
minus all of this stuff. That's going to
cancel with that. That's going to
cancel with that. We're just going
to keep canceling. And we're just going to
have this first term here which is to the 0-th power. Or 1 over k to the 0 power
is going to be equal to 1. Or that this whole sum is going
to be equal to 1 over 1 minus 1 over k. And so we can rewrite s sub k. So we could rewrite s sub k. s sub k is equal to k minus
1 over k factorial times, before I write it up here, let's
see if we can simplify this. If I want to simplify
this I can multiply the numerator and
the denominator by k. If I do that, I have
k on the numerator and then the denominator
becomes k minus 1. So this right here
is k over k minus 1. And these will cancel out. And so this will
be equal to-- we have to be careful
when we cancel it out. Because this kind
of precluded-- well, I'll talk about
that in a second. So this will be equal
to k over k factorial. And what's k over k factorial? Remember, this is equal
to k over-- k factorial is k times k minus 1 times k
minus 2 all the way down to 1. So these k's are
going to cancel out. So it's going to be equal to
1 over k minus 1 times k minus 2 times k minus 3,
all the way down to 1. Which is the same thing
as k minus 1 factorial. Now we want to make sure that we
get the right value for s sub 1 and s sub 2. So let's just get this. So s sub 1, what's that
going to be equal to? And 0 factorial is
always a bit of a debate that people can have. But it becomes a
little bit simpler when you look at the
original expression here. If you put k is equal to 1
here, this just becomes a 0. We lose that because
it cancels later on, and that's why I was kind
of warning to be careful. But we have 1 minus
1 that's clearly going to be 0 regardless
of whether you consider 0 factorial be 1 or 0. It's clearly, clearly
going to be 0 over here. So s sub 1 is going to be 0
times a bunch of other stuff. So s sub 1 is going to be 0. What's s sub 2 going to be? s sub 2 is going to be--
Now we can just use this. It's 1 over 2 factorial which
is just going to be equal to 1. So with that out of
the way, let's think about this sum over here. Let's think about this sum. We have the sum. Actually, before I write it,
it's from k equals 1 to 100. But let's just actually
figure out so when k equals 1, s sub 1 is 0. So this doesn't even
contribute to it. When k is equal to
2, we have a 2 here. Well, let me just
solve it this way since we already
got the 2 there. And then we can start from 3. When k is equal to 2,
s sub 2 is equal to 1. And then we have 2 squared. So that's 4 minus 6 plus 1. So that's negative 1 times
1 which is negative 1. But then we take the
absolute value of it. So the value when-- for
the second term here, the first term is going to be 0. The second term here is
just going to be positive 1. So this sum over here is
going to be equal to 0. This is when k is
equal to 1 plus 1. This is when k is
equal to 2 plus, and now we can worry
about the sum from k is equal to 3 to 100. Notice, it was k
equals 1 to 100. But I did the first two terms. They were a 0 and a 1. So now we have to just
worry about 3 to 100. And let's rewrite
this expression inside the absolute value sign. So inside the
absolute value sign, we have this thing
over here times s sub k. s sub k
is this over here. It's k minus 1 factorial. So you have that whole thing
being divided essentially by k minus 1 factorial. This is the s sub k. It's 1 over k minus 1 factorial. I'm just putting
in the denominator. And you have this thing. You have that thing
in the numerator. Now my first gut instinct
to see if I could factor this, no easy way
to factor this. But my second gut
instinct, hopefully, is to see is if there's any
way that I could express this in terms of k minus 1's
or k minus k's or anything like that, that somehow
will make an easy pattern. Because what tends to be the
case with these type of-- you know you have this crazy sum. Clearly you're not going to
be able to do it by hand. Things are going to cancel out. Normally, you have things
like maybe they oscillate, maybe you have stuff like 1
plus 2 minus 2 plus 3 minus 3. And so all of the
middle terms cancel out, and you just have the first
term and the last term. That tends to be the case for
things like this when there's not some easy
simplification over here or some easy formula like
an infinite geometric sum. So let's see if we can do that. One way we could express this. So let me write this
over here separately. k squared minus 3k plus 1. We could write this as k
squared minus 2k plus 1 minus k. Obviously minus 2k
minus k is negative 3k. And the reason why I do that is
this becomes a perfect square. This is k minus 1 squared. k minus 1 squared minus k. And this looks interesting
because it's now expressing it in terms of simple
terms, especially when you're dealing with
factorials, you like dealing with k minus
1's and k's and maybe it'll simplify a little bit. So I'll write this
expression here as k minus 1 squared minus k. And let's try to simplify it. Let's try to simplify
it even more. So this whole thing is
going to become, or at least we're just dealing with
the sum part right now, we're not really dealing
with the 100 squared over 100 factorial just yet. But this sum is going to
be 1 plus the sum from k is equal to 3 to 100 of
the absolute value of-- we have k minus 1 squared
over k minus 1 factorial. Let me just do it
separately here so I don't have to rewrite this
sigma notation over and over. This expression right over
here is the exact same thing as k minus 1 squared over
k minus 1 factorial minus k over k minus 1 factorial. It's the exact same
thing, but what's interesting about this term,
remember, k minus 1 factorial, that's k minus 1
times k minus-- sorry, k minus-- let me
write it this way. k minus 1 factorial is k
minus 1 times k minus 2 all the way down to 1. That's what this term
is right over here. So we have a k minus
1 squared up here. We have a k minus 1
in the denominator. This will cancel out with one
of the k minus 1's out here. This first term over here
simplifies to k minus 1 in the numerator over-- we
don't have k minus 1 factorial anymore. We now have k minus 2 times
k minus 3 all the way down. So now this is k
minus 2 factorial. And this is still minus k
over k minus 1 factorial. And this looks interesting now. This looks pretty interesting. It looks like a
pattern might emerge. If whatever the k's,
you go in the numerator down one, on the
numerator and down two on the denominator
for the factorial, and then you subtract it. But then for the
next k, it looks like you'll have some cancel. Let's try it out
with some numbers. So this is going to be equal to. When k is equal to
3, this expression is going to be the absolute
value of 2 over 1 factorial. 2 over 1 factorial minus
3 over 2 factorial. So that's when k is equal to 3. And to that we're going to
add when k is equal to 4. When k is equal to 4, we're
going to have 4 minus 1 is 3 over 2 factorial. 3 over 2 factorial minus
4 over 3 factorial. I could keep going. Let's do it when
k is equal to 5. When k is equal to 5, I have 4
over 3 factorial minus 5 over 4 factorial. And I think you see a pattern. We're going to keep
doing this all the way to what's the
absolute last term? The absolute last term
is k is equal to 100. So we go all the way
to k is equal to 100. At 100, you're going to have
the absolute value of 99 over 98 factorial minus
100 over 99 factorial. Now the pattern you
might-- and of course, we have this 1 out front. Now what's this
going to simplify to? You might see the pattern. We're subtracting a
3 over 2 factorial, we're adding a 3
over 2 factorial, we're subtracting a
4 over 3 factorial, then we're adding it. And you might say, well, what
about the absolute values? But when you look over
here, this the second term of these absolute
value signs is always going to be a smaller
value than the first term. And that's because we have
a larger value over here. We only have k
minus 1 factorial. And factorials
increase quite quickly. We only have k minus
1 factorial over here. We only have k minus
2 factorial over here. So this is always going
to be less than this. This right here is 3/2. This right here is 2. This right here is, what is it? 4 over 6. So this is only 2/3. This is 1 and 1/2. So the second term is always
less than the first term. So this is going
to just evaluate to be the same thing without
the absolute value signs. This is always going
to be a positive value. So you actually could ignore
the absolute value signs. And now it becomes I think
pretty pretty pretty clear how the cancellation
is going to happen. You can ignore the
absolute value signs. And now we have
a situation where the second term from this one
for k is equal to 3 cancels with the first term
from k equals 4. Those cancel. Then these two cancel. Then this is going to cancel
with the one from k equals 6. Going to keep happening. This is going to cancel with the
second term from k equals 99. And we're just left
with the one out front. 1 plus this 2 over 1
factorial which is just 2. And then we have minus
100 over 99 factorial. And remember, we also
had this whole time I didn't pay attention to this
100 squared over 100 factorial. But I can do it now. But what is 100 squared? 100 squared is 100 times
100 over 100 factorial. So that's 100 times 99 times
98 all the way down to 1. These cancel out. This is the same thing
as 100 over 99 factorial, which we're going to add to it. So this is 100 over 99
factorial right over here. So let's add that. We have 100 over 99 factorial. Lucky for us, that first term
and this last term cancel out. And we are just left with
1 plus 2 is equal to 3. So that fairly daunting
thing, the main trick was trying to see how to
simplify this in a way that maybe a pattern
would emerge. And that's really the
hard part of this problem. But we get this entire
sum is equal to 3.