Help me find the formula please!

**clarebear14** · February 23rd 2008, 04:53 PM

The question is...

1st term: -1
2nd term: 5
3rd term: 13
4th term: 23
5th term: 25
6th term: 35
7th term: 49

so what could nth term be?

**TheEmptySet** · February 23rd 2008, 05:26 PM

Originally Posted by clarebear14

The question is...

1st term: -1
2nd term: 5
3rd term: 13
4th term: 23
5th term: 25
6th term: 35
7th term: 49

so what could nth term be?

$a_n=n^2+3n-5$ will generate terms 1-4 and 6,7
ie
$a_1=-1$
$a_2=5$
$a_3=13$
$a_4=23$
$a_5=35$
$a_6=49$

and so forth...

This isn't quite the same as yours, are you sure that
$a_5=25$ ?

I hope this helps....

**JaneBennet** · February 23rd 2008, 05:29 PM

Clearly the nth term is

$\frac{-(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)}{720}$

$+\ \frac{5(n-1)(n-3)(n-4)(n-5)(n-6)(n-7)}{-120}$

$+\ \frac{13(n-1)(n-2)(n-4)(n-5)(n-6)(n-7)}{48}$

$+\ \frac{23(n-1)(n-2)(n-3)(n-5)(n-6)(n-7)}{-36}$

$+\ \frac{25(n-1)(n-2)(n-3)(n-4)(n-6)(n-7)}{48}$

$+\ \frac{35(n-1)(n-2)(n-3)(n-4)(n-5)(n-7)}{-120}$

$+\ \frac{49(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)}{720}$

**xifentoozlerix** · February 23rd 2008, 05:31 PM

lol. lagrange ftw!

edit: btw, the fourth term's denominator should be -36 not -9.

$f(n)=\frac{{41}} {{360}}n^6 - \frac{{323}} {{120}}n^5 + \frac{{1789}} {{72}}n^4 - \frac{{907}} {8}n^3 + \frac{{47887}} {{180}}n^2 - \frac{{4604}} {{15}}n + 133 $

**CaptainBlack** · February 24th 2008, 12:40 AM

Originally Posted by clarebear14

The question is...

1st term: -1
2nd term: 5
3rd term: 13
4th term: 23
5th term: 25
6th term: 35
7th term: 49

so what could nth term be?

Anything that you want.

Miss Bennet has put an interpolating polynomial through your seven points,
she could have in fact added an eighth point with value, say 7176, and put
a polynomial throught these eight points.

She could also have added:

$(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)f(x)$

where $f(x)$ is any function on $\mathbb{N}$ to her polynomial and it would still interpolate
your data exactly.

Without some background information constraining solutions, such puzzles have no unique solution.

RonL

**angel.white** · February 24th 2008, 01:25 AM

Originally Posted by JaneBennet

Clearly the nth term is

$\frac{-(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)}{720}$

$+\ \frac{5(x-1)(x-3)(x-4)(x-5)(x-6)(x-7)}{-120}$

$+\ \frac{13(x-1)(x-2)(x-4)(x-5)(x-6)(x-7)}{48}$

$+\ \frac{23(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)}{-9}$

$+\ \frac{25(x-1)(x-2)(x-3)(x-4)(x-6)(x-7)}{48}$

$+\ \frac{35(x-1)(x-2)(x-3)(x-4)(x-5)(x-7)}{-120}$

$+\ \frac{49(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)}{720}$

Clearly.

**JaneBennet** · March 13th 2008, 10:34 PM

Originally Posted by clarebear14

The question is...

1st term: -1
2nd term: 5
3rd term: 13
4th term: 23
5th term: 25
6th term: 35
7th term: 49

so what could nth term be?

Just to clarify, the formula I provided above is on the assumption that the terms of your sequence satisfy a polynomial equation of degree at most 6.

In general, if the first k terms of a sequence $(a_n)_{n=1}^{\infty}$ are given, and it is assumed that the terms of the sequence satisfy a polynomial equation of degree at most $k-1$ , then the nth term of the sequence is

$a_n\ =\ \sum_{j=1}^k{\left(a_j\prod_{1\leq i\leq k,\,i\ne j}{\frac{n-i}{j-i}}\right)}$

Alternatively, you could also let $a_n=\sum_{j=1}^k{c_jn^{k-j}}$ and solve for the coefficients $c_j$ by substituting the first $k$ given values – but this is going to take a hell of a much longer time.

**CaptainBlack** · March 14th 2008, 07:31 AM

Originally Posted by JaneBennet

Just to clarify, the formula I provided above is on the assumption that the terms of your sequence satisfy a polynomial equation of degree at most 6.

In general, if the first k terms of a sequence $(a_n)_{n=1}^{\infty}$ are given, and it is assumed that the terms of the sequence satisfy a polynomial equation of degree at most $k-1$ , then the nth term of the sequence is

$a_n\ =\ \sum_{j=1}^k{\left(a_j\prod_{1\leq i\leq k,\,i\ne j}{\frac{n-i}{j-i}}\right)}$

Alternatively, you could also let $a_n=\sum_{j=1}^k{c_jn^{k-j}}$ and solve for the coefficients $c_j$ by substituting the first $k$ given values – but this is going to take a hell of a much longer time.

This is the Lagrange interpolating polynomial. There is also a method due to Newton that uses the difference table to construct the interpolating polynomial. See here

RonL

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