Let [tex]a_n[/tex] denote the given sequence. [tex]a_n[/tex] has forward differences
{9 - 1, 36 - 9, 100 - 36, ...} = {8, 27, 64, ...} = {2^3, 3^3, 4^3, ...}
If we call the sequence of forward differences [tex]b_n[/tex], then for [tex]n\ge1[/tex],
[tex]b_n=(n+1)^3[/tex]
[tex]b_n[/tex] is defined in terms of [tex]a_n[/tex] for all [tex]n\ge1[/tex] by
[tex]b_n=a_{n+1}-a_n[/tex]
and so [tex]a_n[/tex] is defined recursively by
[tex]a_n=\begin{cases}a_1=1\\a_{n+1}=a_n+(n+1)^3&\text{for }n\ge1\end{cases}[/tex]
We can deduce a pattern for the general [tex]n[/tex]-th term:
[tex]a_2=a_1+2^3[/tex]
[tex]a_3=a_2+3^3=a_1+\displaystyle\sum_{i=1}^2(i+1)^3[/tex]
[tex]a_4=a_3+4^3=a_1+\displaystyle\sum_{i=1}^3(i+1)^3[/tex]
and so on, up to
[tex]a_n=a_1+\displaystyle\sum_{i=1}^{n-1}(i+1)^3[/tex]
We can simplify the right hand side a bit, noticing that [tex]a_1=1=1^3[/tex] matches [tex](i+1)^3[/tex] for [tex]i=0[/tex]:
[tex]a_n=\displaystyle\sum_{i=0}^{n-1}(i+1)^3[/tex]
and to simplify things a bit more, we shift the index of summation:
[tex]a_n=\displaystyle\sum_{i=1}^ni^3[/tex]
You should know that the right side has a nice closed form (look up "Faulhaber's formula" if you don't):
[tex]a_n=\dfrac{n^2(n+1)^2}4[/tex]
