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# Sequences, methods of differences Watch

1. Using the method of differences, or otherwise, derive the term for the series

2 + 10 + 30 + 68 + 130 + ...
-8-20-38-62-
-12-18-24
-6-6-6

I understand you have to use the quadratic formula, but I can't get my head on how to plugin the numbers and really just the overall method.
2. (Original post by theonn)
Using the method of differences, or otherwise, derive the term for the series

2 + 10 + 30 + 68 + 130 + ...
-8-20-38-62-
-12-18-24
-6-6-6

I understand you have to use the quadratic formula, but I can't get my head on how to plugin the numbers and really just the overall method.
how does the quadratic formula come into it?

start at the bottom

compare it to the sequence generated by

U = n^3 (as there are three rows before the difference evens out)

note that the bottom row is the same, so is the next row up, but the size of the row corresponding to the linear term is one bigger than it should be.

we'll take a different sequence as an example that of U=n^2-2n+1

so the sequence and differences are as follows

0 1 4 9 16 one greater than n^2 - 2n that we get from below so +1
1 3 5 7 is two less than if no linear term so -2n
2 2 2 tells us there is a single square term

Spoiler:
Show
the answer to yours is n^3 + n?
3. (Original post by theonn)
Using the method of differences, or otherwise, derive the term for the series

2 + 10 + 30 + 68 + 130 + ...
-8-20-38-62-
-12-18-24
-6-6-6

I understand you have to use the quadratic formula, but I can't get my head on how to plugin the numbers and really just the overall method.
You had to take the differences three times before arriving at a constant term, so this is going to be a cubic, rather than a quadratic.

You can just plug in the values to , where [/latex]u_n[/latex] is the n'th term, and then solve. You'll need 4 simultaneous equations.

I have assumed you're interested in what the n'th term is, and not what the sum is.

Edit: See previous poster, for a method more closely connected with that of differences
4. (Original post by ghostwalker)
You had to take the differences three times before arriving at a constant term, so this is going to be a cubic, rather than a quadratic.

You can just plug in the values to , where [/latex]u_n[/latex] is the n'th term, and then solve. You'll need 4 simultaneous equations.

I have assumed you're interested in what the n'th term is, and not what the sum is.
2 + 10 + 30 + 68 + 130 + ...
-8-20-38-62-
-12-18-24
-6-6-6

Is that right? What's next?
5. (Original post by theonn)
2 + 10 + 30 + 68 + 130 + ...
-8-20-38-62-
-12-18-24
-6-6-6

Is that right? What's next?
Yes. Now solve. Putting it into augmented matrix form would be easiest.

But also see the other posters method, which I'm not familiar with.
6. (Original post by ghostwalker)
Yes. Now solve. Putting it into augmented matrix form would be easiest.

But also see the other posters method, which I'm not familiar with.
I have, his one seems much more confusing. How do I go about solving them? haha

Edit: Is it =
7. (Original post by theonn)
I have, his one seems much more confusing. How do I go about solving them? haha

Edit: Is it =
natninja had the correct solution, and you can always check by substituting different values of n.

I suspect you had the matrix reduction correct, and just went wrong the on the final part.

Should be
8. If you look at the question, you realized that:

n = 1 , 2 , 3 , 4 , 5
U_n= 2, 10, 30, 68, 130

How can we get from

n = 1 --> 2
n = 2 ---> 10
n = 3 ----> 30
n = 4 ----> 68

So forth, we see that we multiply first by (1x2) = 2, then (2x5) = 10, then (3x10) = 30, then (4x17) = 68.

If you like at the multiplicative factors, you see for U_1 = x2, U_2 = x5, U_3 = x10, U_4 = x17...

2,5,10,17 is simply the sequence (n^2 + 1)

As we're multiplying n by the multiplicative factor, we have

Of course, you could just look at the sequence and realize that it approximately

Then you'll see the difference of each term is simply .

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