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What is the relation between the general formula of the sum of $n$ terms of an arithmetic progression, $a n^2+b n+c$, and the first term $a+b+c$ and the common difference?

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@mixedmath I think he means the formula for the sum of the arithmetic progression, not the formula for the arithmetic progression – Thomas Andrews Jul 17 '12 at 16:38
@ThomasAndrews: That makes more sense. I'll delete my answer. – Cameron Buie Jul 17 '12 at 17:32

Suppose that the terms we are adding up are $a_1,a_2,a_3$, and so on. Let $s_1=a_1$, $s_2=a_1+a_2$, $s_3=a_1+a_2+a_3$, and so on.

If $a_1,a_2,a_3,\dots$ is an arithmetic sequence, let $d$ be the common difference. So $a_1=a_1$, $a_2=a_1+d$, $a_3=a_1+2d$, and so on. We therefore have $a_k=a_1+(k-1)d$. It follows that $$s_n=a_1+(a_1+d)+ (a_1+2d)+\cdots +(a_1+(n-1)d).$$ The above sum is equal to $$na_1+d(1+2+\cdots +(n-1)).$$ By a standard result, $1+2+\cdots+(n-1)=\frac{1}{2}n(n-1)$. Thus $$s_n=na_1+\frac{1}{2}(n)(n-1)d=\frac{d}{2}n^2 +\left(a_1-\frac{d}{2}\right)n.$$ Comparing with the proposed formula $s_n=an^2+bn+c$, we get that $c=0$ and $d=2a$.

Conversely, suppose that $c=0$, and the sum of the first $n$ terms is given by the formula $s_n=an^2+bn$. The $k$-th term $a_k$ is then $(ak^2+bk)-(a(k-1)^2+b(k-1))$. This simplifies to $a_k=2ak +b$. Write $d$ instead of $2a$. Then $a_k=kd+b=b-d+(k-1)d$, exactly the expression for the $k$-th term of an arithmetic sequence with first term $b-d$ and common difference $d$.

Another way: We are given that $s_n=an^2+bn+c$. We will show that the sequence $a_1,a_2,a_3,\dots$ is an arithmetic sequence iff $c=0$, and will determine the common difference.

The first term (where we take first to mean $n=1$) is $a+b+c$. So $a_1=a+b+c$.

The sum $s_2$ of the first $2$ terms is $4a+2b+c$. So $a_2$ must be $(4a+2b+c)-(a+b+c)$, that is, $3a+b$. Thus the difference $a_2-a_1$ between the second term and the first must be $(3a+b)-(a+b+c)$, which is $2a-c$. This is the common difference, if the sequence really is an arithmetic sequence. Let's check whether it really is.

Let $k\ge 2$. We have $s_k=ak^2+bk+c$, and $s_{k-1}=a(k-1)^2+b(k-1)+c$. Thus $$a_k=s_k-s_{k-1}=(ak^2+bk+c)-(a(k-1)^2+b(k-1)+c)=2ak-a+b.$$ It follows that $a_{k+1}=2a(k+1)-a+b$. The difference between $a_{k+1}-a_k$ between the $(k+1)$-th term and the $k$-th term is therefore $2a$ when $k \ge 2$.

We saw that the difference $a_2-a_1$ is $2a-c$, and that all further differences are $2a$. We therefore have an arithmetic progression iff $c=0$. Then the common difference is $2a$.

Remark: If we are willing to take it on faith that the expression $an^2+bn+c$ does give the sum of an arithmetic sequence, say with first term $w$ and common difference $d$, then we can operate more concretely. Perhaps the following is intended.

We have $a+b+c=w$, $4a+2b+c=w+(w+d)=2w+d$, and $9a+3b+c=w+(w+d)+(w+2d)=3w+3d$. Use the first two equations to eliminate $c$, also the last two equations. We get $3a+b=w+d$ and $5a+b=w+2d$. Subtract to eliminate $b$. We get $2a=d$. From $3a+b=w+d$ we then get $w=3a-2a+b=a+b$. Then from $a+b+c=w$ we get $c=0$.

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If the sum of $n$ terms of an arithmetic sequence is $$ s(n)=\color{#C00000}{an^2+bn+c}\tag{1} $$ then $c$ is the sum of $0$ terms; that is, $$ c=s(0)=0\tag{2} $$ The general term of the arithmetic sequence would be $$ \begin{align} a(n) &=s(n)-s(n-1)\\ &=(b-a)+(2a)n\tag{3} \end{align} $$

The first term of the arithmetic sequence would be $$ a(1)=\color{#C00000}{a+b}\tag{4} $$ Thus, the common difference of the arithmetic sequence is $$ \begin{align} a(n)-a(n-1) &=((b-a)+(2a)n)-((b-a)+(2a)(n-1))\\ &=\color{#C00000}{2a}\tag{5} \end{align} $$ As for a relation among $(1)$, $(4)$, and $(5)$, we get $$ \color{#C00000}{an^2+bn+c}=n(\color{#C00000}{a+b})+\frac{n^2-n}{2}\color{#C00000}{2a}\tag{6} $$

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