Question: Consider $\Phi$ to be the space of sequences that have finitely many non-zero terms.
The space is not closed in $\ell\_1$, therefore $\ell\_1/\Phi$ with the quotient topology is not Hausdorff, and so it cannot be metrizable. However, does there exist a metric on $\ell\_1/\Phi$ that gives rise to a non-trivial topology. And even stronger question is, is $\ell\_1/\Phi$ normable?

Answer: 
We first look at the simpler expression
$$\sum\_{k=1}^n k\cdot k!.$$
This is 
$$1\cdot 1!+2\cdot 2! +3\cdot 3!+\cdots +n \cdot n!.$$
In general, $k \cdot k!=(k+1)\cdot k!-k!=(k+1)!-k!$.
It follows that the sum above is equal to 
$$(1!-0!)+(2!-1!)+(3!-2!) +(4!-3!)+\cdots +((n+1)!-n!).$$
Add up. There is a whole lot of cancellation, and we get $(n+1)!-1$.

Now we turn to our problem, which can be rewritten as 
$$\sum\_{k=1}^n(k+1)^2\cdot k! +\sum\_{k=1}^n k \cdot k!,$$
since $k^2+3k+1=(k+1)^2+k$. We have already computed the second sum. The first sum is 
$$\sum\_{k=1}^n (k+1)\cdot (k+1)!.$$
The same collapsing argument then shows that this sum is 
$$(n+2)! -2.$$
Or else we can recycle the previous result, noting that we are dealing with $\sum\_{j=2}^{n+1}j\cdot j!$, which is $1$ less than $\sum\_{j=1}^{n+1}j\cdot j!$.
Finally, add up. We get
$$[(n+2)(n+1)!-2] +[(n+1)!-1].$$
This is $(n+3)(n+1)!-3$.

**Comment**: But the above solution actually does not answer the question! The OP asked that induction not be used. However, induction *was* used, albeit in a subtle hidden way. We *saw* the systematic cancellation, it was obvious. But a "proper" complete proof would use the cancellation up to the $k$-th term to prove cancellation up to the $(k+1)$-th term. Much of the time when one sees ellipses ($\dots$) in a mathematical expression, induction is, technically speaking, needed to fill in the full formal details. Not that this should make any practical difference in our mathematical behaviour: Obvious is still obvious.
I would use $$\sum\_{i=1} ^n (i+2)! - i! = (n+2)!+(n+1)! - 2! - 1 !$$