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

Guest Jul 23, 2014

Best Answer 

 #2
avatar+93633 
+10

 

 

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

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.

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

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

Melody  Jul 23, 2014
 #1
avatar+20024 
+5

        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  Jul 23, 2014
 #2
avatar+93633 
+10
Best Answer

 

 

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

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.

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

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

Melody  Jul 23, 2014
 #3
avatar+93633 
0

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

Melody  Jul 23, 2014
 #4
avatar+20024 
+5

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:

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

heureka  Jul 23, 2014
 #5
avatar+93633 
0

Thanks Heureka

Melody  Jul 23, 2014

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