$v_0+v_1+.....+v_n=\frac{1}{3}(2n^2-11n-13)$ I used the sum of the arithmetic sequences in simple steps

Question

I extract in two different ways the base $r$ and the first term $v_0$ of the arithmetic sequence $v_n$, knowing that for each natural number $n$ it is: $v_0+v_1+.....+v_n=\frac{1}{3}(2n^2-11n-13)$

I used the sum of the arithmetic sequences in simple steps and I found the base and the first term so that $v_0=-\frac{13}{3}$

$r=\frac{4}{3}$

Who can benefit me in another way

Answer 1

$v_0=-\frac{13}{3}.$

Also, for any $n\geq1$ we obtain: $$v_n=\frac{1}{3}(2n^2-11n-13)-\frac{1}{3}(2(n-1)^2-11(n-1)-13)=\frac{4}{3}n-\frac{13}{3}.$$ Now, we see that even for $n=0$ the last formula is valid and it's enough to check that $v$ is indeed an arithmetic progression.

We see that $$v_{n+1}-v_n=\frac{4}{3}$$ and we are done!

Answer 2

$$3v_m=2(m^2-(m-1)^2)-11(m-(m-1))=\cdots=4m-13$$

So, $$r=v_n-v_{n-1}=\dfrac{4(n-(n-1))}3=?$$

Answer 3

In general, given $v_0+\cdots+v_n=an^2+bn+c$, we have $v_0=c$. Since $a+b+c=v_0+v_1=2v_0+r=2c+r$ we get $r=a+b-c$.

Answer 4

We may also view the sum-formula in this way. If we put $ \ v_0 + v_1 + \ \ldots \ + v_n \ = \ \frac{2n^2 \ - \ 11n \ - \ 13}{3} \ \ , $ into correspondence with the familiar expression $ \ n·v_0 \ + \ \frac{n·(n - 1)}{2}·r \ = \ \frac{r}{2}·n^2 \ + \ (v_0 - r)·n \ \ $ (using your notation), we note that the term-difference $ \ r \ = \ \frac43 \ \ $ can be "read-off" immediately from the quadratic term. But we also see that the constant term $ \ -\frac{13}{3} \ $ cannot be accounted for, which suggests that the first term $ \ v_0 \ $ is somehow "appended" to the expression for the given sum-formula. The fact that the quadratic polynomial can be factored as $ \ 2n^2 \ - \ 11n \ - \ 13 \ = \ (2n - 13)·(n + 1) \ \ $ is suspicious: the formula seems to be starting the sum from $ \ v_1 \ \ . $

If we "shift" the index of the conventional expression "upward" by $ \ 1 \ \ , $ we obtain $$ \ (n+1)·v_0 \ + \ \frac{(n+1)·n}{2}·r \ = \ \frac{r}{2}·n^2 \ + \ \left(v_0 + \frac{r}{2} \right)·n \ + \ v_0 \ \ . $$ It is now clear that $ \ \mathbf{r \ = \ \frac43} \ $ and $ \ \mathbf{v_0 \ = \ -\frac{13}{3}} \ \ ; $ we confirm this by checking the linear term, $ \ \left(v_0 + \frac{r}{2} \right)·n \ = \ \left(-\frac{13}{3} + \frac23 \right)·n \ = \ -\frac{11}{3}·n \ \ . $