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September 30th, 2017, 02:44 PM   #1
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How to determine the nth term in this series?

I have this series, but it does not seem to follow a constant difference, or at least is not implicit

$\displaystyle -1,0,5,14,27,...$

How can I determine the nth term, let's say the 16th term and the sum up to that term? Is there an easy way without resorting to induction?

Last edited by skipjack; September 30th, 2017 at 07:50 PM.
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September 30th, 2017, 09:24 PM   #2
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The nth term is 2n² - 5n + 2. (This is the simplest formula I could spot.)
Hence the sum of the first n terms is (4n³ - 9n² - n)/6.

The first 16 terms are -1, 0, 5, 14, 27, 44, 65, 90, 119, 152, 189, 230, 275, 324, 377, 434.

Successive differences are 1, 5, 9, 13, 17, 21, 25, 29, . . .
Successive second differences are 4, 4, 4, 4, 4, 4, 4, . . .
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September 30th, 2017, 10:33 PM   #3
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Originally Posted by skipjack View Post
The nth term is 2n² - 5n + 2. (This is the simplest formula I could spot.)
Hence the sum of the first n terms is (4n³ - 9n² - n)/6.

The first 16 terms are -1, 0, 5, 14, 27, 44, 65, 90, 119, 152, 189, 230, 275, 324, 377, 434.

Successive differences are 1, 5, 9, 13, 17, 21, 25, 29, . . .
Successive second differences are 4, 4, 4, 4, 4, 4, 4, . . .
I tried, I mean really hard to guess which recursive formula could be used in this series. But is there any kind of algorithm that can be used by following steps on how to attack these situations? I mean the equation you proposed works, but it looks as if some kind of magician trick like taking a rabbit out of a hat (if you know what I mean).

The second part which involves the sum of terms in the sequence seems kind of logical as there exists these useful formulas (which can be obtained from any book).

$\displaystyle \sum_{k=1}^{n}k^{3}=\left (\frac{n(n+1)}{2} \right )^{2}$

$\displaystyle \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6} $

Last edited by skipjack; September 30th, 2017 at 11:06 PM.
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September 30th, 2017, 11:14 PM   #4
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As the second differences are all 4, one can find a quadratic formula of the form 4n²/2 + bn + c for the nth term. It's now easy to determine the values of b and c.
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September 30th, 2017, 11:20 PM   #5
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I can't speak for how skipjack came up with the recursion, but you can always try to solve for the coefficients of a polynomial.

Suppose you have $N$ points

let $p(n)=\displaystyle \sum_{k=0}^{N-1}~c_k n^k$

and you end up with a system of $N$ equations when you plug in values, e.g. in this case

$-1 = p(1),~0=p(2), 5=p(3), 14=p(4), 27=p(5)$

Solving this, you find $c_0 = 2,~c_1=5,~c_2=2,~c_3=0,~c_4=0$

Last edited by skipjack; September 30th, 2017 at 11:27 PM.
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October 3rd, 2017, 06:39 PM   #6
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I can't speak for how skipjack came up with the recursion, but you can always try to solve for the coefficients of a polynomial.

Suppose you have $N$ points

let $p(n)=\displaystyle \sum_{k=0}^{N-1}~c_k n^k$

and you end up with a system of $N$ equations when you plug in values, e.g. in this case

$-1 = p(1),~0=p(2), 5=p(3), 14=p(4), 27=p(5)$

Solving this, you find $c_0 = 2,~c_1=5,~c_2=2,~c_3=0,~c_4=0$
That's probably the key in solving this problem. However, I find it rather a bit of non-explicit in how to translate the sum proposed into the terms, or constants shown.

Can you show the example of how to use the sum to solve for, let's say, $c_1$? Sorry if I do ask it this way, but it is not too obvious for the casual learner.

Last edited by skipjack; October 4th, 2017 at 11:56 AM.
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October 4th, 2017, 04:21 AM   #7
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Of course, a "sequence" doesn't have to follow any simple rule at all. When I was a freshman in college, my math professor gave an example of the sequence "15, 16, 17, 18, 31, 32, 33, 53, 54".

Those were the numbers of the subway stations on his way home after work. His particular train shunted off one "line" to another twice.

The best you can do with a problem like this is look for the "simplest" formula. And, of course, what is "simplest" may depend upon the person setting the problem.

Last edited by skipjack; October 4th, 2017 at 11:58 AM.
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October 4th, 2017, 12:20 PM   #8
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Using the formula 4n²/2 + bn + c,
putting n = 1 leads to -1 = 2 + b + c, and putting n = 2 leads to 0 = 8 + 2b + c.
Subtracting the first equation from the second gives 1 = 6 + b, so b = -5.
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October 4th, 2017, 01:11 PM   #9
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$\{-1,0,5,14,27,...\}$

$\{-1 \cdot 1, \, 0\cdot 3, \, 1 \cdot 5, \, 2 \cdot 7, \, 3\cdot 9, \, ... , \, (n-2)(2n-1), \, ... \}$
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October 4th, 2017, 04:05 PM   #10
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I originally noticed that the nth term is $T_{2\text{n}-3}$ - 1 ≡ (2n - 3)(2n - 2)/2 - 1 ≡ 2n² - 5n + 2.
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