if 1=2 2=10 3=30 4=68 then 5=?

$$\begin{tabular}{|c|c|c|c|c|c|} \hline $x$&1&2&3&4&5\\\hline $f(x)$&2&10&30&68&\\\hline \end{tabular}$$

Okay they are not getting bigger by a set multiplication factor - as the x gets bigger, f(x) gets MUCH bigger.

Try squared, no still not big enough -- Try cubed.  That looks close.  Add x and you've got it.

$$\begin{tabular}{|c|c|c|c|c|c|} \hline $x$&1&2&3&4&5\\\hline $f(x)$&2&10&30&68&\\\hline $x^2$&1&4&9&16&\\\hline $x^3$&1&8&27&64&\\\hline $x^3+x$&2&10&30&68&\\\hline \end{tabular}$$

$$\\f(x)=x^3+x\\\\ f(5)=5^3+5=125+5=130$$

        if     1  =  2

                2 = 10

                3 = 30

                4 = 68

        then 5 =   ?

$$\begin{array}{ccccc} n+1& p(n)& \Delta_1 & \Delta_2 & \Delta_3 \\ \\ 1&2& \\ &&8\\ 2&10&&12 \\ &&20&&\textcolor[rgb]{1,0,0}{6}\\ 3&30&&18&\downarrow \\ &&38&&(\textcolor[rgb]{1,0,0}{6})\\ 4&68&&(24) \\ &&(62)\\ 5&(130)& \end{array}$$

 5 =   130

Polynom

$$p(n) = n^3+3*n^2+4*n+2 \quad | \quad n = 0,1,2,3, ...$$

Heureka, you have got the same answer as me but I don not understand what you have done.  

Hi Melody,

this is a arithmetic succession. Order is 3 ( So i hold 6 as a constant number in Delta 3 so the order is 3 ). I got the Polynom from Internet.

Beginning bei n= 0. If i set x = n+1 so the Polygon can written as:

$$p(x) \\ = (x-1)^3+3(x-1)^2+4(x-1)+2 \\ = x^3-3x^2+3x-1+3(x^2-2x+1)+4x-4+2 \\ = x^3-3x^2+3x-1+3x^2-6x+3+4x-4+2\\ = x^3-3x^2+3x^2+3x-6x+4x -1+3-4+2\\ = x^3+x \\ p(x)=x^3+x \quad | \quad x=1,2,3,... \quad Your\quad solution$$

Thanks Heureka