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## Integrated math 3

# Describing numerical relationships with polynomial identities

Explore numerical relationships in polynomial identities, demonstrating how they can reveal patterns in sequences of integer squares. Learn how algebra can prove that the difference between successive terms in a sequence of integer squares is always an increasing odd number.

## Want to join the conversation?

- I think I need some help to understand what is going on. As I understood the difference between any number by it's previous number is 2n+1. Then, let's take 9.

but the difference between 9 and 4 is not 19. do you guys have any idea?`2(9) + 1 = 19`

(5 votes)- The polynomial '2n+1' is defined as the
**difference**between any number by it's previous number. You have the right idea; however, we're looking at what is inbetween. To clarify:

The sequence is n², and you want to put 9.

9² = 81

Then,

10²=100

The goal of '2n+1' is the difference between (in this example), 10²-9² = 100-81 =**19**, your answer.

Another example is that you want to plug in 13 and 14 for n², and you want to find the difference:

13² = 169

14² = 196,

Using our formula,

2(13)+1 = a 27 number distance. Let's now check:

14²-13²=27(18 votes)

- How would you use this in real life? Economics maybe?(3 votes)
- I'm no expert, but I have heard that sequences (such as the ones shown in this video) show up in computer programming. They also show up in calculus and statistics, which is used in real life for sure. Also, algebra as a subject overall improves your critical thinking skills, which is important to have in life.(20 votes)

- On the second example, how do you know that n is n^2?(4 votes)
- Because the second sequence is describing the numbers in the first sequence squared. n is the value of any number in the first sequence, so n^2 is the value of any number in the first sequence squared (which is what the second sequence describes).(2 votes)

- Please correct me if I am wrong.

Since Sal has given us the differences between two square number(2n+1), we can figure the sequences of the two series.

First series(n+1):

**a(0)=0**

{

a(n)=a(n-1) +1*a(n-1) as previous term**n-1 as n minus 1*

e.g:

a(1)=1

a(2)=a(2-1)+1

=1+1=2

Second series(2n+1):`*a(0)=0`

{

a(n)=a(n-1) 2(n-1)*

e.g:

a(2)=4

a(3)=a(3-1)+2(3-1)+1

=4+2(2)+1=9

Any numbers can be used with this sequence.(2 votes)- Hi, I just read what you wrote.

I'm just wondering if the first series is necessary? If you simplify it a bit more, it becomes a(n) = n. So is there really a point in this case?(3 votes)

- How is 0^2 0?

Because if you think about it, then

0^2 = 0

0*0 = 0

And if we divide 0 from both sides, we get

0 = 0/0

But any number divided by 0 is undefined, which means

0 = undefined

Which is not true(1 vote)- The problem occurred when you said “if we divide zero from both sides.”(4 votes)

- This was a bit of a "non-sequitur" so far.(3 votes)
- I think in real life, someone will just throw the second sequence right on your face and said find some pattern for me!

to be honest, if I encounter such circumstance, I may not find the n^2 pattern(3 votes) - I need a "like" button for this video!(2 votes)
- at3:10I noticed it was a +1 following the n^2 + 2n but shouldn't it be +2?(1 vote)
- No, 1^2=1 therefore it should be n^2+2n+1.(2 votes)

- would this not be 2n-1?(1 vote)
- You're on the right track, but the equation solves for the
*difference*between`n^2`

and the squared number**after**`n`

. You must be thinking about the*difference*between`n^2`

and the squared number**before**`n`

.

For example, if`n=5`

, then we can use the equation`2n+1`

to find the difference between`5^2`

and`6^2`

(notice`6^2`

is the number that comes after`5^2`

).

Of course, we can always solve this the lengthy way:`5^2=25 6^2=36`

36-25=11

But the equation solves this a lot faster:`2n+1 = 2(5)+1 = 10+1 = 11`

The equation`2n+1`

is like a shortcut for solving the difference in between.

——————————————————————————————————————

If you are having a hard time understanding how we got from`(n+1)^2–n^2`

to`2n+1`

, here's how:`(n+1)^2`

is a perfect square, and when we expand it, we get`n^2+2n+1`

. Then we just add in the remaining`–n^2`

so we get:`n^2+2n+1–n^2`

Notice we have both a positive and a negative`n^2`

in the equation, so those cancel out. So we are left with:`2n+1`

(2 votes)

## Video transcript

- [Instructor] What we're going
to do in this video is use what we know about polynomials
and how to manipulate them and what we've talked about
of whether two polynomials are equal to each other for all values of the variable that they're written in, so whether we're dealing
with a polynomial identity. And we're going to use
those skills in order to prove some properties of
relationships between numbers. So if I were to list out some integers, I could go zero, I could
go one, I could go two, three, four, five. And if I were to list
out the squares of these, if I were to create a
sequence of integer squares, well, zero squared would be zero, one squared would be one, two squared is four, three squared is nine, four squared is 16, five squared is 25. And we could, of course,
keep going in either case. But the first thing I
want you to think about, before you even write down a polynomial or try to construct one, is look at this sequence
of integer squares. And do you see any pattern in terms of the difference
between successive terms of this sequence of integer squares? All right, now let's
think about this together. So to go from zero to one, you add one. And to go from one to four, you add three. To go from four to nine, you add five. To go from nine to 16, you add seven. It seems like a pattern here. As we go to successive terms of this sequence of integer squares, we're adding increasing odd numbers. So I'm guessing that if I add nine here, which is the next odd
number, I'm gonna get to 25, and that indeed is the case. And you could test that out. Well, what, if I add 11, which would be the next odd
number, what do I get to? I get to 36, which is the square of six. But how can we feel good
that this always is true, that this never breaks down? Well, one way to do it is to think a little bit more generally, and that's where our
algebra and our knowledge of polynomials are going to be useful. So let's say we go all the way, and we're just speaking generally now. So we have the number n, and
then with the next number after that is going to be n plus one. And then if we think about
what the corresponding terms in the sequence of
integer squares would be, well, that would be, when we square it, when we get to n, we would get n squared. And when we get to n plus one, we would have n plus one squared. And let's see if we could
think about what the difference between these two things are. The difference between 25 and 16 is nine. Difference between 16 and nine is seven. So let's think about what the difference between n plus one
squared and n squared is. And how do we write that as a polynomial? Well, it'll just be n plus one squared minus n squared. And now let's see if we can rewrite this, algebraically manipulate
this so we can set up a polynomial identity that
describes this pattern that we just saw. So what I'll do is I'm
just going to expand out n plus one squared right over there. So that is going to be n
squared plus two n plus one. And then we have this
minus n squared here, so minus n squared. And so we see that n squared
minus n squared cancels out. And so we can rewrite
everything we have here as n plus one squared minus n squared. So this is really the difference
between successive terms in our sequence of integer squares is going to be equal to two n plus one for any integer n. Well, for any integer n, what
is two n plus one going to be? And especially here, we're dealing with the positive integers. Well, for any integer n, this
is going to be an odd integer. If you take any integer,
you multiply it by two, this part is going to be even. But then you add one to that, you're going to get an odd integer. And you can also see that this increases by two as n increases. So when you go from one odd integer, you go add two to the next odd integer. You add two to the next odd integer, which is exactly what is described there. So this is pretty neat. We've just used a little bit of algebra, a little bit of what we know
about polynomial identities to show that the difference
between successive terms in this sequence of integer
squares right over here is going to be increasing odd numbers.