<s>[INST] How to find $n$'th term of the sequence $3, 7, 12, 18, 25, \ldots$? $$3, 7, 12, 18, 25, \ldots$$
This sequence appears in my son's math homework. The question is to find the $n$'th term. What is the formula and how do you derive it?
 [/INST] for the series: 3, 7, 12, 18, 25, …
$3=0+3\\7=3+4\\12=7+5\\18=12+6\\25=18+7$
We can find that: each term equals previous term plus n+2
$t_0=0$
$t_1=t_0+(1+2)=(1+2)$
$t_2=t_1+(2+2)=(1+2)+(2+2)$
$t_3=t_2+(3+2)=(1+2)+(2+2)+(3+2)$
...
$t_n=t_{n-1}+(n+2)$
$=(\color{red}1+\color{green}2)+(\color{red}2+\color{green}2)+(\color{red}3+\color{green}2)+...+(\color{red}n+\color{green}2)$
$=\color{red}{(1+...+n)}+\color{green}{(2n)}$
$=\frac{n(n+1)}{2} + 2n$
$=\frac{(n^2+5n)}{2}$
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