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?(Crying)

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- Feb 23rd 2008, 04:53 PMclarebear14Help me find the formula please!
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?(Crying) - Feb 23rd 2008, 05:26 PMTheEmptySet
$\displaystyle a_n=n^2+3n-5$ will generate terms 1-4 and 6,7

ie

$\displaystyle a_1=-1$

$\displaystyle a_2=5$

$\displaystyle a_3=13$

$\displaystyle a_4=23$

$\displaystyle a_5=35$

$\displaystyle a_6=49$

and so forth...

(Thinking)

This isn't quite the same as yours, are you sure that

$\displaystyle a_5=25$?

I hope this helps.... - Feb 23rd 2008, 05:29 PMJaneBennet
Clearly the

*n*th term is

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

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

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

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

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

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

$\displaystyle +\ \frac{49(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)}{720}$ - Feb 23rd 2008, 05:31 PMxifentoozlerix
lol. lagrange ftw!

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

$\displaystyle 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

$ - Feb 24th 2008, 12:40 AMCaptainBlack
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:

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

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

your data exactly.

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

RonL - Feb 24th 2008, 01:25 AMangel.white
- Mar 13th 2008, 10:34 PMJaneBennet
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 $\displaystyle (a_n)_{n=1}^{\infty}$ are given, and it is assumed that the terms of the sequence satisfy a polynomial equation of degree at most $\displaystyle k-1$, then the*n*th term of the sequence is

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

Alternatively, you could also let $\displaystyle a_n=\sum_{j=1}^k{c_jn^{k-j}}$ and solve for the coefficients $\displaystyle c_j$ by substituting the first $\displaystyle k$ given values – but this is going to take a hell of a much longer time. (Tongueout) - Mar 14th 2008, 07:31 AMCaptainBlack
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