The start of a quadratic sequence is
\[
6, \quad 15, \quad 28, \quad 45, \quad 66, \quad \ldots
\]
What is the $n^{\text {th }}$ term rule for this sequence?

Step 1 ：First, find the differences between consecutive terms: $15-6=9$, $28-15=13$, $45-28=17$, $66-45=21$

Step 2 ：Next, find the differences between the differences: $13-9=4$, $17-13=4$, $21-17=4$

Step 3 ：Since the second differences are constant, the sequence is quadratic. Let the nth term be $an^2+bn+c$

Step 4 ：Plug in the first term: $a(1)^2+b(1)+c=6$

Step 5 ：Plug in the second term: $a(2)^2+b(2)+c=15$

Step 6 ：Plug in the third term: $a(3)^2+b(3)+c=28$

Step 7 ：Solve the system of equations: $\begin{cases} a+b+c=6 \\ 4a+2b+c=15 \\ 9a+3b+c=28 \end{cases}$

Step 8 ：Subtract the first equation from the second and third equations to eliminate c: $\begin{cases} 3a+b=9 \\ 8a+2b=22 \end{cases}$

Step 9 ：Multiply the first equation by 2 and subtract it from the second equation to eliminate b: $2a=4$

Step 10 ：Solve for a: $a=2$

Step 11 ：Substitute a back into the first equation to find b: $3(2)+b=9$, so $b=3$

Step 12 ：Substitute a and b back into the first equation to find c: $2(1)^2+3(1)+c=6$, so $c=1$

Step 13 ：The nth term rule is $an^2+bn+c=2n^2+3n+1$

Step 14 ：\(\boxed{2n^2+3n+1}\)