The sequence
1, 3, 7, 13, 21, ...
has first-order differences
2, 4, 6, 8, ...
Let [tex]a_n[/tex] denote the original sequence, and [tex]b_n[/tex] the sequence of first-order differences. It's quite clear that
[tex]b_n=2n[/tex]
for [tex]n\ge1[/tex]. By definition of first-order differences, we have
[tex]b_n=a_{n+1}-a_n[/tex]
for [tex]n\ge1[/tex], or
[tex]a_{n+1}=a_n+2n[/tex]
By substitution, we have
[tex]a_n=a_{n-1}+2(n-1)[/tex]
[tex]\implies a_{n+1}=(a_{n-1}+2(n-1))+2n[/tex]
[tex]\implies a_{n+1}=a_{n-1}+2(n+(n-1))[/tex]
[tex]a_{n-1}=a_{n-2}+2(n-2)[/tex]
[tex]\implies a_{n+1}=(a_{n-2}+2(n-2))+2(n+(n-1))[/tex]
[tex]\implies a_{n+1}=a_{n-2}+2(n+(n-1)+(n-2))[/tex]
and so on, down to
[tex]a_{n+1}=a_1+2(n+(n-1)+\cdots+2+1)[/tex]
You should know that
[tex]1+2+\cdots+(n-1)+n=\dfrac{n(n+1)}2[/tex]
and we're given [tex]a_1=1[/tex], so
[tex]a_{n+1}=1+n(n+1)=n^2+n+1[/tex]
or
[tex]a_n=(n-1)^2+(n-1)+1\implies\boxed{a_n=n^2-n+1}[/tex]
Alternatively, since we already know the sequence is supposed to be quadratic, we can look for coefficients [tex]a,b,c[/tex] such that
[tex]a_n=an^2+bn+c[/tex]
We have
[tex]a_1=a+b+c=1[/tex]
[tex]a_2=4a+2b+c=3[/tex]
[tex]a_3=9a+3b+c=7[/tex]
and we can solve this system for the 3 unknowns to find [tex]a=1,b=-1,c=1[/tex].
