What type of sequence is this?

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franz

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I tried proving this sequence = the formula (2n)(n+1)
n = 1 2 3 4
6 20 42 72
multiple ways. Initially using Arithmetic sequences with a second difference, and Geometric sequences which just didn't make much sense. Now i'm just completely out of ways to prove it.
I already know the formula which achieves these numbers, (2n)(n+1), but i cannot prove coherently that this is the correct solution. My question being, how do I solve this, and how do I prove it works?
 
franz said:
I already know the formula which achieves these numbers, (2n)(n+1), but i cannot prove coherently that this is the correct solution. My question being, how do I solve this, and how do I prove it works?
Click to expand...
Yes, there is a mistake in the formula.
In any case, it is possible to find the formula by looking at the second differences.
Post your work if you want it checked.
 
franz said:
I tried proving this sequence = the formula (2n)(n+1)
n = 1 2 3 4
6 20 42 72
multiple ways. Initially using Arithmetic sequences with a second difference, and Geometric sequences which just didn't make much sense. Now i'm just completely out of ways to prove it.
I already know the formula which achieves these numbers, (2n)(n+1), but i cannot prove coherently that this is the correct solution. My question being, how do I solve this, and how do I prove it works?
Click to expand...
How do you "know" that is the formula? Did you at least check whether it works for the terms shown? It doesn't, so it can't be proved! And if you checked all four terms and it worked, then no more proof would be needed.

You presumably meant (2n)(2n+1), which I could see just by observing that the numbers are 2*3, 4*5, 6*7, 8*9.

And that is all the proof that is possible! If a formula fits the numbers given, then it is a valid answer (even if your teacher had a different formula in mind). There are, in fact, infinitely many possible formulas.
 
Dr.Peterson said:
How do you "know" that is the formula? Did you at least check whether it works for the terms shown? It doesn't, so it can't be proved! And if you checked all four terms and it worked, then no more proof would be needed.

You presumably meant (2n)(2n+1), which I could see just by observing that the numbers are 2*3, 4*5, 6*7, 8*9.

And that is all the proof that is possible! If a formula fits the numbers given, then it is a valid answer (even if your teacher had a different formula in mind). There are, in fact, infinitely many possible formulas.
Click to expand...
Yes, I meant (2n)(2n+1). Thank you for the help!
 
Given any n pairs of numbers there exist a polynomial of degree at most n-1 that will give those pairs,

Here you have four pairs so there exist a polynomial of degree three or less that will give them.

We can write the polynomial as \(\displaystyle ax^3+ bx^2+cx+ d\). We have to determine 4 values, a, b, c, and d, and need 4 equations to do this.

When x= 1, y= 6 so a+ b+ c+ d= 6.
When x= 2, y= 20 so 8a+ 4b+ 2c+ d= 20
When x= 3, y= 42 so 27a+ 9b+ 3c+ d= 42
When x= 4, y= 72 so 64a+ 16b+ 4c+ d=72

Each equation has "+ d" so we can eliminate d just by subtracting.
The second minus the first gives 7a+ 3b+ c= 14.
The third minus the second gives 19a+ 5b+ c= 22.
The fourth minus the third gives 37a+ 7b+ c= 30.

Now we have "+ c" in each equation so:
The second minus the first gives 12a+ 2b= 8 or 6a+ b= 4.
The third minus the second gives 18a+ 2b= 8 or 9a+ b= 4.

Subtracting one more time, 3a= 0 so a= 0. Yes, these points are given by a polynomial of degree 2 or less.
6a+ b= b= 4.
7a+ 3b+ c= 12+ c= 14 so c= 2.
a+ b+ c+ d= 4+ 2+ d= 6+ d= 6 so d= 0.

These four (x, y) pairs are given by \(\displaystyle y= 4x^2+ 2x= 2x(2x+ 1)\) or, using n instead of x,

2n(2n+ 1).

Check:
When n= 1, 2n(2n+1)= 2(3)= 6.
When n= 2, 2n(2n+ 1)= 4(5)= 20.
When n= 3, 2n(2n+ 1)= 6(7)= 42.
When n= 4, 2n(2n+ 1)= 8(9)= 72.
 
Last edited:
Actually there was a silly typo which I have corrected. The formula is 2n(2n+1), not 2n(n+1).
I should have done the check before!
 
inc
HallsofIvy said:
Given any n pairs of numbers there exist a polynomial of degree at most n-1 that will give those pairs,

Here you have four pairs so there exist a polynomial of degree three or less that will give them.

We can write the polynomial as \(\displaystyle ax^3+ bx^2+cx+ d\). We have to determine 4 values, a, b, c, and d, and need 4 equations to do this.

When x= 1, y= 6 so a+ b+ c+ d= 6.
When x= 2, y= 20 so 8a+ 4b+ 2c+ d= 20
When x= 3, y= 42 so 27a+ 9b+ 3c+ d= 42
When x= 4, y= 72 so 64a+ 16b+ 4c+ d=72

Each equation has "+ d" so we can eliminate d just by subtracting.
The second minus the first gives 7a+ 3b+ c= 14.
The third minus the second gives 19a+ 5b+ c= 22.
The fourth minus the third gives 37a+ 7b+ c= 30.

Now we have "+ c" in each equation so:
The second minus the first gives 12a+ 2b= 8 or 6a+ b= 4.
The third minus the second gives 18a+ 2b= 8 or 9a+ b= 4.

Subtracting one more time, 3a= 0 so a= 0. Yes, these points are given by a polynomial of degree 2 or less.
6a+ b= b= 4.
7a+ 3b+ c= 12+ c= 14 so c= 2.
a+ b+ c+ d= 4+ 2+ d= 6+ d= 6 so d= 0.

These four (x, y) pairs are given by \(\displaystyle y= 4x^2+ 2x= 2x(2x+ 1)\) or, using n instead of x,

2n(2n+ 1).

Check:
When n= 1, 2n(2n+1)= 2(3)= 6.
When n= 2, 2n(2n+ 1)= 4(5)= 20.
When n= 3, 2n(2n+ 1)= 6(7)= 42.
When n= 4, 2n(2n+ 1)= 8(9)= 72.
Click to expand...
incredible useful! helped tons, thank you.
 
For these types of problems - where the method is not specified -I invoke Excel,
plot the points and ask for trend-line equation
 
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