Help with quadratic sequence for son's homework I'm sorry this is really basic. I'm terrible with math and I'm struggling to help my son with homework:
Find an expression in terms of n for the nth term in this sequence:
$0 , 9 , 22 , 39, 60, \ldots$
We can get the $2n^{2}$ part of the answer but just can't get the rest. Any help gratefully appreciated.
Sorry again for there basic question.
 A: Given that the sequence is quadratic, start by assuming that
$$a_n=An^2+Bn+C$$
for some $A,B,C$. Since $a_0=0$, it follows immediately that $C=0$ and
$$a_n=An^2+Bn$$
Furthermore, since $a_1=9$ and $a_2=22$, you have
$$a_1=A+B=9$$
$$a_2=4A+2B=22$$
and you have the system of equations
$$A+B=9$$
$$4A+2B=22$$
When solved, this yields $A=2$ and $B=7$. Thus you have
$$a_n=2n^2+7n$$
A: Every quadratic sequence has successive differences in arithmetic progression.  That is to say, if the original terms are $$a_0, a_1, a_2, a_3, \ldots,$$ then the sequence defined by $$b_k = a_{k+1} - a_k, \quad k = 0, 1, 2, \ldots$$ is an arithmetic progression, which in turn implies $$c_k = b_{k+1} - b_k$$ is a constant sequence.
Specifically, we find $$b_0 = 9, \quad b_1 = 22 - 9 = 13, \quad b_2 = 39-22 = 17, \quad b_3 = 60 - 39 = 21.$$  Then $$c_0 = 13 - 9 = 4, \quad c_1 = 17 - 13 = 4, \quad c_2 = 21 - 17 = 4.$$  Thus the formula for $b_k$ is $$b_k = b_0 + c_0 k = 9 + 4k.$$  How do we use this to find $a_k$?  We observe $a_{k+1} = a_k + b_k = a_k + 9 + 4k$.  We unfold this recursion once to give $$a_{k+1} = (9 + 4k) + (9 + 4(k-1)) + a_{k-2} = 18 + 4(k + (k-1)).$$  If we unfold this all the way, we get $$a_{k+1} = 9(k+1) + 4(k + (k-1) + \cdots + 1) + a_0,$$ which gives $$a_k = 9k + 2k(k-1) + a_0 = 2k^2 + 7k + a_0.$$  Since $a_0 = 0$, we get $$a_k = k(2k+7).$$

If we reason in reverse, we can establish the above property of quadratic sequences (namely, their successive differences are in arithmetic progression).  For suppose $$a_k = Ak^2 + Bk + C$$ for $k = 0, 1, 2, \ldots$.  Then $$b_k = a_{k+1} - a_k = A(k+1)^2 + B(k+1) + C - Ak^2 - Bk - C = 2Ak + (A+B).$$  This is an arithmetic sequence with common difference $2A$.  Thus we can use this rather than the unfolding we did above:  Since the common difference was found to be $4$, we know $2A = 4$ or $A = 2$.  Since the initial term (offset) of $\{b_k\}$ is $b_0 = 9$, and this is $A+B$, we find $B = 9 - 2 = 7$.  Finally, since $a_0 = 0$, we get $C = 0$, and $$a_k = 2k^2 + 7k + 0 = k(2k+7).$$
A: Another way to get the answer is to see what you have left.
If you suspect that you have the $2n^2$ part, then write that out:
$$0, 2, 8, 18, 32$$
Subtract this sequence from what you have to get
$$(0-0), (9-2), (22-8), (39-18), (60-32) \\ \to 0, 7, 14, 21, 28$$
Now it's fairly clear that the missing part is $7n$.
