$$\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$&1&2&3&4&5\\\hline
$f(x)$&2&10&30&68&\\\hline
\end{tabular}$$
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
$$p(n) = n^3+3*n^2+4*n+2 \quad | \quad n = 0,1,2,3, ...$$
$$\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$&1&2&3&4&5\\\hline
$f(x)$&2&10&30&68&\\\hline
\end{tabular}$$
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.
$$\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$&1&2&3&4&5\\\hline
$f(x)$&2&10&30&68&\\\hline
$x^2$&1&4&9&16&\\\hline
$x^3$&1&8&27&64&\\\hline
$x^3+x$&2&10&30&68&\\\hline
\end{tabular}$$
$$\\f(x)=x^3+x\\\\
f(5)=5^3+5=125+5=130$$
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:
$$p(x) \\
= (x-1)^3+3(x-1)^2+4(x-1)+2 \\
= x^3-3x^2+3x-1+3(x^2-2x+1)+4x-4+2 \\
= x^3-3x^2+3x-1+3x^2-6x+3+4x-4+2\\
= x^3-3x^2+3x^2+3x-6x+4x -1+3-4+2\\
= x^3+x \\
p(x)=x^3+x \quad | \quad x=1,2,3,... \quad Your\quad solution$$