# Need help is hard

1. Jun 28, 2007

### alasdair101

okay i think this goes in the right section, but is hard, what is the equation to predict the next numbers?

when 1=2, 2=4, what is the equation, to predict the next numbers

x---y
1---2
2---4
3---8
4---12
5---18
6---24
7---32
8---40
9---50
10---60
11---72
12---84
13---98
14---112
15---128
16---144
17---162
18---180
19---200
20---220

and same for this one too
x---y
1---21
2---47
3---84
4---131
5---189
6---257
7---336
8---425
9---525
10---635
11---756
12---887
13---1029
14---1181
15---1344
16---1517
17---1701
18---1895
19---2100
20---2315

Last edited: Jun 29, 2007
2. Jun 29, 2007

### matt grime

Please don't use an equals sign when you don't mean an equality. 1 is not equal to 2, nor 2 to 4.

The first is 'straight-foward', though you probably want an answer in some unreasonably nice form. Just look at the differences of consecutive terms: 2,4,4,6,6,8,8,10,10, etc. This means: if n=2k, or 2k+1 for some integer k, than the n'th term is.....

3. Jun 29, 2007

### HallsofIvy

Staff Emeritus
You understand, I hope, that there are an infinite number of equations, f(n), that will give precisely those numbers up to n= 20 and then give different numbers for n= 21?

For s sequence of n given numbers, there exist a unique n-1 degree polynomial which gives that sequence. That means there must exist 19 degree polynomials giving your two sequences. On way to find them is to Write out the general 19 degree polynomial, substitute the given values for x and y and solve the resulting 20 equations. Perhaps simpler is the Lagrange polynomial:

For each (xi,yi) pair for the product
$$y_i \frac{(x- x_1)(x-x_2)\cdot\cdot\cdot(x-x_n)}{(x_i-x_1)(x_i-x_2)/cdot/cdot/cdot(x_i-x_n)}$$
where the prolducts in the numerator and denominator include all x values except xi