We're given the sequence,
[tex]\{b_n\}_{n=3}^\infty=\{25,39,55,73,\ldots\}[/tex]
Since [tex]b_n[/tex] is quadratic, [tex]a_n[/tex] should also be quadratic. Replace [tex]n[/tex] with [tex]n-2[/tex] (if [tex]n=3[/tex], then [tex]n-2=1[/tex]) in the definition of [tex]b_n[/tex], and let [tex]a,b,c[/tex] denote the new coefficients:
[tex]n^2+7n-5=a(n-2)^2+b(n-2)+c[/tex]
Expand the right side:
[tex]n^2+7n-5=an^2+(-4a+b)n+(4a-2b+c)[/tex]
Coefficients of terms with the same degree should be the same:
[tex]\begin{cases}a=1\\-4a+b=7\\4a-2b+c=-5\end{cases}\implies a=1,b=11,c=13[/tex]
So the new sequence is the same, with
[tex]\{a_n\}_{n=1}^\infty=\{25,39,55,73,\ldots\}[/tex]
