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.
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.