Week 6 / Sequences W2025
Week 6
We go to Chapter 4, section 1. Sequences… !? But you have seen this! Com’on! You
survived Cal 2! That is nothing… ! Let us “review” a bit of sequences.
A sequence is an ordered list of numbers. (an)n>0 or 1…
There are the known ones, the squares, the cubes, the “triangular numbers”, the sum of
integers up to n. We know this sequence, also the triangle numbers: n(n+1)/2. We have the
powers of 2, the Fibonacci sequence,
See the excel file. Walk through. Identify each one, to keep in memory to identify
variation of the same theme.
What is new: the recursive sequences… We are simply defining them, we’ll come back
with some tools to “solve” them very soon. See the two ways of defining a sequence: a(n)
simply as a function of n and a(n) as a function of a(n-1) and/or a(n-2) …
The Hanoi Towers minimal number of moves to solve a 2 disks, a 3 disks… and see if you
can recursively see a solution… If you add another, 4th, disk… what do you need to do?
See the Investigate! Preview.
Do example 4.1.8 3, 4, 5. You have to be patient… Try the others tonight!
You know the sigma symbol and how we create a sequence of partial sums: 𝑏𝑛 = ∑𝑛𝑖=1 𝑎𝑖
A sum can start at 0… or 1… or at any number! See Example 4.1.10 3.
And have a try with the 1, 5, 17, 53, 161, 485…
We can tackle it via its sequence of differences… and observe what is going on…
Do you want to learn Python or stick with Java? How would you write the recursive
function an = 3(an-1)+2, a0 = 1, and print the 10 first numbers of the sequence?
Do the reading questions 1, 2.
Do the practice problems 1 to 5. Additional ones, 9, 10, 11.
Moving to 4.2 : arithmetic and geometric sequences… Ok, you know these two.
Review!
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Do the Investigate… ! with the dots… 😊 The middle one is … fun…!
Arithmetic : an = a0 +n·d. The d is the difference between two successive terms. The first
term, a0, is key as well as the d.
If you want the nth term, you have it: a0 +n·d. Or similarly, in the recursive manner, an =
an-1 + d.
If you want the sum of an arithmetic sequence, you can “flip it” like Gauss, and have Sn =
∑𝑛𝑖=0 𝑎𝑖 = (n-1)a0+n(n+1)d/2.
Sum Recursive: Sn+1 = Sn + (n+1)d + a0.
Do the example 4.2.1.
Geometric sequences: not adding but multiplying. We have a0 and a1 is equal to a0r and a2
= a1r or a0r2. Note that our formulas work if you work from n=0!
So we have an = a0rn. If you want a recursive definition, an+1 = anr. Again, we are interested
into the partial sum of a geometric sequence, or series. Sn = ∑𝑛𝑖=0 𝑎𝑖 . The trick: you
multiply by r and subtract. You obtain : Sn = a0(1-rn+1)/(1-r).
A second look at the triangular sequence…
Do example 4.2.2. Reading question 1.
Suggested exercises : 1, 2, 3, 4, 5, 9. Additional exxercises : 4, 8.
We move to the section 4.3: finite differences.
You recognize an arithmetic sequence when from an to an+1, there is a constant difference.
That is, an+1 - an = d, constant, for all n. We found the close formula for this : an = a0+nd.
This is a linear expression of n.
What if I need to do twice the difference, that is, ∆2? Example :
1, 4, 9, 16, 25… I do the difference : 3,5,7, 9… ooooohhh… Another difference :
2, 2, 2, 2… Stop !!! This is a constant difference! But I am at the second difference level..
So the close formula will be a quadratic of n…
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Week 6 / Sequences W2025
Ok, yes, easy, n2. Do example 4.3.3 as is and with Polynomial fitting. Highlight how to
find a0.
Do example 4.3.6 and use Excel to do diff… 😊
And as usual… we all know “Garbage In, Garbage Out” motto. Be careful, if you have
enough terms of your sequence, you can fit a polynomial… but make sure it works outside
of the terms provided… check a bit after… It is worth it!
Exercise : Reading question 1, practice problems 1, 2, 3, 4, 5, 7, 8. Additional questions
#12
And now we tackle 4.4 Exponential sequences.
Let us do the preview activity 4.4.1. Easy… see the excel sheet. For the last question…
what if you look at the two first sequences we have seen … 2^n and 3^n … ???
Dealing with partial sums of a geometric series. Example 4.4.1. Can we generalize this
result? Use a0 and r instead of 3 and 2…
Well… if we apply what we know to a linear second degree recursive definition of a series
and want to find a closed expression… We will use the Characteristic Root Technic. The
secret is… we seem to suspect a geometric nature behind exponential sequences… Let us
see. We want to solve a second-degree linear recursive sequence. Example : an = 6an-1
+7an-2. We assume there is a r that fits the job of being the ratio: an = rn. But what does it
mean? If r works, it means rn = 6rn-1 + 7rn-2. Or put all terms on one side : rn - 6rn-1 - 7rn-2
=0. Well well… put rn-2 up front : rn-2 (r2 - 6r - 7) =0. Oh! Quadratic to solve! We find two
values for r : -1 and 7. So our solution will be a mixture of (-1)n and (7)n. Note that the
solution depends on the two first values! We are given that a0 = 1 and a1 = 3. Ok, how do
we find the particular solution? We need two values, we have them! We solve this 2 by 2:
X(-1)^0 + y(7)^0 = 1 and
X(-1) + y(7) = 3. That is, x+y = 1 and 7y-x = 3. Plug x = 1-y into 7y –(1-y) = 3. We find
y=1/2 and x=1/2 as well. So, for our example, ½[(-1)n + (7)n] = an. Verify! You got the
twist. Practice with example 4.4.4. What if the characteristic
The golden ratio and Fibonacci.
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What if the characteristic polynomial has a single r squared like (x-r)^2? We do not have
two roots… so an =arn+bnrn.
Let the students do example 4.4.5. Suggested exercises : reading question 1, practice
problems #1, 2, 3, 4, 5. Additional exercises 3 and 4. And moving along!
Moving to Induction, 4.5. The domino principle.
Give 4 minutes to students to do the investigate! About the unit digit of 6 n. How to prove
this obvious fact?
You want to prove that the sum of the squares ∑𝑛𝑖=1 𝑖 2 is n(n+1)(2n+1)/6.
The first thing is to check if it works for n=1, n=2… Let us see.
N=1 : 1(2)(3)/6 = 1, yes.
N=2 : 2(3)5/6 = 5 and yes = 1 + 4.
Ok, it works for 1 and 2. You should check also for n=3… please!
Can we “prove” it works for 3, 4, 5, 6 … ? Mostly for the “…” part!?
Here is the strategy behind a proof by induction.
- We formalize an equality, a property: We propose that ∑𝑛𝑖=1 𝑖 2 is n(n+1)(2n+1)/6.
- We test for an early value: it works for n = 1, 2.
- We put forward the following implication: If ( ∑𝑘𝑖=1 𝑖 2 = k(k+1)(2k+1)/6 for a fixed k
2
value ) then ( ∑𝑘+1
𝑖=1 𝑖 = (k+1)(k+2)(2(k+1)+1)/6 ) must be true.
- Careful, what we prove is not the first part, that ∑𝑘𝑖=1 𝑖 2 = k(k+1)(2k+1)/6 is true. What
we prove is the implication! If the first part IS TRUE, then the second part IS ALSO
TRUE.
How do we prove this implication? As we do with any implication, assume the first part,
work hard until you get a proof of the second part.
Let us do the proof of the example we just wrote.
We assume that for a value k, ∑𝑘𝑖=1 𝑖 2 = k(k+1)(2k+1)/6 is true.
If we add (k+1)2 on both sides:
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Week 6 / Sequences W2025
∑𝑘𝑖=1 𝑖 2 + (k + 1)2 = k(k+1)(2k+1)/6 +(k+1)2 The first side is simply… the sum up to
k+1.
2
2
∑𝑘+1
𝑖=1 𝑖 = k(k+1)(2k+1)/6 +(k+1) =
2
∑𝑘+1
𝑖=1 𝑖 =
(𝑘+1)[ 𝑘(2𝑘+1)+6(𝑘+1)]
6
=
2
But wait… We have now ∑𝑘+1
𝑖=1 𝑖 =
𝑘(𝑘+1)(2𝑘+1)+6(𝑘+1)2
6
(𝑘+1)[2𝑘 2 +𝑘+6𝑘+6]
6
=
=
(𝑘+1)[ 𝑘(2𝑘+1)+6(𝑘+1)]
6
(𝑘+1)[2𝑘 2 +7𝑘+6]
6
=
(𝑘+1)[2𝑘+3][𝑘+2]
6
(𝑘+1)[(𝑘+1)+1][2(𝑘+1)+1]
6
2
What we proved is that IF ∑𝑘𝑖=1 𝑖 2 = k(k+1)(2k+1)/6 is true, THEN ∑𝑘+1
𝑖=1 𝑖 =
(𝑘+1)[(𝑘+1)+1][2(𝑘+1)+1]
6
is also true, for whatever k value.
Because we know it works for n = 1, we proved that if it works for 1, it MUST work for 2,
and if it works for 2, it MUST work for 3, … etc… ad vitam aeternam.
In short, the first domino felt. We checked it. And we proved that if one domino falls, the
next one is close enough to fall as well. So the whole chain of dominos will fall…
Two steps: 1st the verification for a given small n; 2nd the proof of the implication { if it
works for n, then it must work for n+1 }. And bingo!
Look at n2 < 2n for all n > 5. A bit challenging. We use twice the fact of n2 < 2n.and the
fact that 4n < n squared, n>5!
Difference between induction and strong induction: a very subtle change in the
assumption. The first one assumes that p(k) is true imply p(k+1) is true, for all k.
The second one assumes that p(k) for any k<n is true, imply p(n) is true, for all n.
That is… very subtle! Fine, now you know… Strong induction is asking a bit more but
gives you a more solid proof. It can hurt in the game of breaking a problem in two smaller
problems, not just n-1 type of problems. See example 4.5.5.
Suggested problems of 4.5: reading problem 1, practice problems #1, 2, 3, 4, 5, 6.
Additional problems 2, 3, 4, 14, 15, 16.
Ouf! We stop here for chapter 4! Moving to the next subject: generating functions! For
fun… Chapter 6.2.
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