Question 874586
The differences between consecutive terms are
9, 15, 30, 54, 87, 129.
If they were the same, it would be an arithmetic sequence,
and calling the initial term {{{x[0]}}} , and the common difference {{{d}}} ,
term number {{{n}}} would be
{{{x[n]=x[0]+n*d}}} .
In other words, {{{x[n]}}} would be a linear function of {{{n}}} .
 
The differences of the differences are
{{{15-9=6}}}
{{{30-15=15}}}
{{{54-30=24}}}
{{{87-54=33}}}
{{{129-87=42}}}
If those differences were the same, {{{x[n]}}} would be a quadratic function of {{{n}}} .
 
The differences of the differences of the differences are all {{{9}}} :
{{{15-6=9}}}
{{{24-15=9}}}
{{{33-24=9}}}
{{{42-33=9}}}
{{{x[n]}}} must be a cubic function of {{{n}}} .
 
From the differences, we can make a table and figure out that the next terms are
505, 745, 1054.
The table would be
{{{matrix(10,3,1,difference,second,10,9,difference,
25,15,6,55,30,15,109,54,24,196,87,33,325,129,42,
325+180=505,129+51=180,42+9=51,
505+240=745,180+60=240,51+9=60,
745+309=1054,240+69=309,60+9=69)}}}
 
Finding a formula is a little harder.
The fact that the third differences are constant means that {{{x[n]}}} is a cubic function of {{{n}}}
{{{x[n]=a*n^3+b*n^2+c*n+d}}}
{{{x[0]=d=1}}}
{{{x[1]=a+b+c+1=10}}}-->{{{a+b+c=9}}}
{{{x[2]=a*2^3+b*2^2+c*2+1=25}}}-->{{{8a+4b+2c+1=25}}}-->{{{8a+4b+2c=24}}}-->{{{4a+2b+c=12}}}
{{{x[3]=a*3^3+b*3^2+c*3+1=55}}}-->{{{27a+9b+3c+1=55}}}-->{{{27a+9b+3c=54}}}-->{{{9a+3b+c=18}}}
Solving {{{system(a+b+c=9,4a+2b+c=12,9a+3b+c=18)}}} we find can find the formula easily enough:
From {{{system(a+b+c=9,4a+2b+c=12,9a+3b+c=18)}}} ,
subtracting the first equation from the other two we get
{{{system(a+b+c=9,3a+b=3,8a+2b=9)}}}
Subtracting 2 times {{{3a+b=3}}} , or {{{6a+2b=6}}} , from {{{8a+2b=9}}} we get
{{{system(a+b+c=9,3a+b=3,2a=3)}}}-->{{{system(a+b+c=9,3*1.5+b=3,a=1.5)}}}-->{{{system(a+b+c=9,4.5+b=3,a=1.5)}}}-->{{{system(a+b+c=9,b=-1.5,a=1.5)}}}-->{{{system(a=9,b=-1.5,a=1.5)}}}
The formula is {{{highlight(x[n]=1.5n^3-1.5n^2+9n+1)}}}
We already know that it will give us
{{{x[0]=1}}} , {{{x[1]=10}}} , {{{x[2]=25}}} , and {{{x[3]=55}}} ,
because we used those terms to find the coefficients.
Using the formula found for the next given terms, we see that it works: 
{{{x[4]=1.5*4^3-1.5*4^2+9*4+1=1.5*64-1.5*16+36+1=96-24+36+1=109}}}
{{{x[5]=1.5*5^3-1.5*5^2+9*5+1=1.5*125-1.5*25+45+1=1.5(125-25)+46=150+46=196}}}
{{{x[6]=1.5*6^3-1.5*6^2+9*6+1=1.5*216-1.5*36+54+1=324-54+54+1=325}}}