In general - How do I sum polynomial sequences like $3.7 + 5.9+...$ in closed form? I'm perusing an old book (Operators by D.R. Dickinson) which has a section showing how to sum for example
$$3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \cdots + (2n+1)(2n+3)(2n+5)$$
For the generic, closed-form solution the book gives, it matters that the difference between successive factors in a given term is the same as the difference between the corresponding factors in successive terms.
Here we have $3 \cdot 5 \cdot 7$ with factor difference of $2$, and also the first term (3.5.7) starts with $3$, the second term (5.7.9) starts with 5.
For the record, what they do to sum the above is to firstly write
$$u_{r} = \prod_{s = 0}^{m-1} [a + (r+s-1)b]$$
This way, you can change either r or s and compensate (by adjusting limit m) to get the same thing.
Now they have an auxillary definition, applying when $r \geq 1$
$v_{r}=\prod_{s = 0}^{m} [a + (r+s-1)b] \quad \quad$
This time, notice index $s$ stops at $m$ rather than (m-1).
for $r \ge 1$
$$v_{r} = u_{r}.[a + (r+m-1)b]$$
also
$$v_{r_{-1}} = \prod_{s = 0}^{m} [a + (r+s-2)b] = [a + (r+s-2)b].u_{r}$$
Therefore
$$v_{r} - u_{r-1} = (m+1).b.u_{r}$$
$$\sum_{r = 1}^{n} u_{r} = \dfrac{1}{(m+1).b}(v_n-v_0)$$
I hope I've written that correctly - I have slightly changed the book's rather opaque presentation.
Okay - But now how about this
$3.7 + 5.9+...+(2n+1)(2n+5)$
The way they tackle this is to first derive some "standard forms" (which I have labeled F1, F2, F3)
$\sum_{r = 1}^{n} 1 = n \quad \quad (F_1)$
$\sum_{r = 1}^{n} r = \dfrac{1}{2}n(n+1) \quad \quad (F_2)$
$\sum_{r = 1}^{n} r(r+1) = \dfrac{1}{3}n(n+1)(n+2) \quad \quad (F_3)$
Now, expand the new problem, i.e.
$u_r = (2r+1)(2r+5) = 4r^2+12r+5 = 4r(r+1) + 8r + 5$
Notice the term $r(r+1)$ is like the LHS of form F3, $8r$ is like F2 and $5$ is like F1.
Therefore
$\sum_{r = 1}^{n} u_r = \dfrac{4}{3}n(n+1) + 4n(n+1) + 5n$
Now the above is some work and is not a closed form in a generic sense for series of similar form having say $6$ factors per term.
 
So - my question is
Is there a more generic way to approach this, not requiring those "Standard Forms" and not asking us to multiply out all factors in $u_r$?
 A: The general term can be written as
$$T_n=(2r+1)(2r+3)(2r+5)$$
$$8T_n=(2r+1)(2r+3)(2r+5)(2r+7)-(2r-1)(2r+1)(2r+3)(2r+5)$$
$$8T_n=V_{n+1}-V_n$$
Where $$V_n=(2r-1)(2r+1)(2r+3)(2r+5)$$
Now $$S=\sum_{r=1}^n T_r=\frac{1}{8}\sum_{r=1}^n(V_{r+1}-V_{r})$$
Now telescope to get $\displaystyle S=\boxed{\frac{V_{n+1}-V_1}{8}}$
A: You can write the given series as \begin{align} S_n = \sum_{r =1}^{n}((2r+1)(2r + 3)(2r + 5)) \\= \sum_{r =1}^{n} 8 \, r^{3} + 36 \, r^{2} + 46 \, r + 15 \\= 8 \sum_{r =1}^{n} r^3 + 36 \sum_{r =1}^{n}r^2 + 46 \sum_{r =1}^{n} r + \sum_{r =1}^{n}15 \end{align}
Using $$ \sum_{r =1}^{n}r^3 = \frac{1}{4} \, {\left(n + 1\right)}^{2} n^{2}$$ and $$\sum_{r =1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$$ and $$\sum_{r =1}^{n}r = \frac{n(n+1)}{2}$$ we find that

$$ S_n = 2 \, n^{4} + 16 \, n^{3} + 43 \, n^{2} + 44 \, n = {\left(2 \, n^{2} + 8 \, n + 11\right)} {\left(n + 4\right)} n$$

A: As the general term is a polynomial, you can expand it and use the Faulhaber formulas for the respective terms. As Faulhaber is itself a closed-form, you will get a closed-form in all cases.
In the given example,
$$8n^3+36n^2+46n+15\to 2n^2(n+1)^2+6n(n+1)(2n+1)+23n(n+1)+15n\\=2n^4+16n^3+43n^2+44n.$$

You can also obtain that polynomial by Lagrangian interpolation on the first n+1 terms of the series.
Newtonian interpolation is even better for general terms that are products of $k$ integers in arithmetic progression.
In your case:
$$\Delta_1=(2n+3)(2n+5)(2n+7)$$
$$\Delta_2=6(2n+5)(2n+7)$$
$$\Delta_3=27(2n+7)$$
$$\Delta_3=54$$
