Let [tex]a_n[/tex] denote the [tex]n[/tex]-th term of the sequence, starting with the index [tex]n=0[/tex]. Denote by [tex]b_n[/tex] the forward differences of [tex]\{a_n\}_{n\ge0}[/tex], that is
[tex]b_n=a_{n+1}-a_n[/tex]
for [tex]n\ge0[/tex]. We have
[tex]\{b_n\}_{n\ge0}=\{4,7,10,13,16,\ldots\}[/tex]
which indicates [tex]b_n[/tex] is an arithmetic sequence with common difference between terms of 3 and starting with 4, so that
[tex]b_n=4+3n[/tex]
for [tex]n\ge0[/tex]. So we have
[tex]a_0=1[/tex]
[tex]a_1=a_0+b_0=1+4[/tex]
[tex]a_2=a_0+b_0+b_1=1+4+7[/tex]
[tex]a_3=a_0+b_0+b_1+b_2=1+4+7+10[/tex]
and so on, with the [tex]n[/tex]th term given by
[tex]a_n=a_0+b_0+b_1+\cdots+b_{n-2}+b_{n-1}[/tex]
[tex]a_n=1+4+7+\cdots+(4+3(n-2))+(4+3(n-1))[/tex]
[tex]a_n=1+4+7+\cdots+(3n-2)+(3n+1)[/tex]
Reversing the order of all but the first term in the sum gives
[tex]a_n=1+(3n+1)+(3n-2)+(3n-5)+\cdots+7+4[/tex]
so that
[tex]2a_n=2+(4+3n+1)+(7+3n-2)+\cdots+(3n-2+7)+(3n+1+4)(3n+1+1)[/tex]
[tex]2a_n=2+\underbrace{(3n+5)+(3n+5)+\cdots+(3n+5)+(3n+5)}_{n\text{ times}}[/tex]
[tex]2a_n=2+n(3n+5)[/tex]
[tex]\implies\boxed{a_n=1+\dfrac{n(3n+5)}2}[/tex]
for [tex]n\ge0[/tex].
