If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Integrated math 3

### Course: Integrated math 3>Unit 2

Lesson 6: Polynomial identities

# 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.
``2(9) + 1 = 19``
but the difference between 9 and 4 is not 19. do you guys have any idea?
• 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
• How would you use this in real life? Economics maybe?
• 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.
• On the second example, how do you know that n is n^2?
• 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).
• 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.
• 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?
• 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.”
• This was a bit of a "non-sequitur" so far.
• 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
• I need a "like" button for this video!
• at I 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.
• 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=3636-25=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`