Question 970169
 {{{1}}},{{{8}}},{{{27}}},{{{58}}}

first find the pattern:

To find the pattern, I will list the numbers, and find the differences for each pair of numbers. That is, I will subtract the numbers in pairs (the first from the second, the second from the third, and so on), like this:

{{{1}}}.........{{{8}}}..........{{{27}}}..........{{{58}}}.....

.....{{{7}}}.........{{{19}}}.........{{{31}}}.... the "first differences", are not all the same value, I'll continue subtracting:
..........{{{12}}}.........{{{12}}}....=> the second differences are the same, so the polynomial for this sequence of values is a quadratic
Since the formula for the terms is a quadratic, then I know that it is of the form:

    {{{an^2 + bn + c }}} for some numbers {{{a}}}, {{{b}}}, and {{{c}}}

Now I have to find those numbers by plugging in some of the values from the sequence, and then solving the resulting system of equations. 

if {{{n=1}}} first term is {{{1}}}

{{{a*1^2 + b*1 + c=1 }}}
{{{a + b + c=1}}} ..........eq.1


if {{{n=2}}} second term is {{{8}}}

{{{a*2^2 + b*2 + c=8 }}}
{{{4a + 2b + c=8}}} ..........eq.2


if {{{n=3}}} third term is {{{27}}}

{{{a*3^2 + b*3 + c=27}}}
{{{9a + 3b + c=27}}} ..........eq.3

start with

{{{4a + 2b + c=8}}} ..........eq.2
{{{a + b + c=1}}} ..........eq.1
--------------------------------subtract  
{{{3a + b=7}}} --------solve for {{{b}}}
{{{b=7-3a}}}.........................(1)

now use 
{{{9a + 3b + c=27}}} ..........eq.3
{{{4a + 2b + c=8}}} ..........eq.2
--------------------------------subtract 
{{{5a+b=19}}}--------solve for {{{b}}}

{{{b=19-5a}}}.........................(2)

use (1) and (2) to find {{{a}}}

{{{7-3a=19-5a}}}

{{{5a-3a=19-7}}}

{{{2a=12}}}

{{{highlight(a=6)}}}

now find {{{b}}}

{{{b=7-3a}}}.........................(1)
{{{b=7-3*6}}}
{{{b=7-18}}}
{{{highlight(b=-11)}}}

go to eq.1 substitute {{{a}}} and {{{b}}} , and find {{{c}}}

{{{a + b + c=1}}} ..........eq.1
{{{6 -11 + c=1 }}}
 {{{-5 + c=1 }}}
{{{c=1+5}}}
{{{highlight(c=6)}}}

so, your polynomial,or nth term formula is:

{{{a[n] = 6n^2-11n+6}}}  .......(for all terms given)

now use formula to find the next two numbers in this series:

{{{a[5] = 6*5^2-11*5+6}}}

{{{a[5] = 150-55+6}}} 

{{{a[5] = 101}}}  


{{{a[6] = 6*6^2-11*6+6}}}

{{{a[6] = 216-66+6}}} 

{{{a[6] = 156}}} 


so, with these terms, your sequence is: {{{1}}},{{{8}}},{{{27}}},{{{58}}},{{{101}}},{{{156}}}