For any natural $n$, how could we prove that $\sum\limits_{i=1} ^n (i^2+3i+1) i!= (n+3)(n+1)! - 3$ How could we prove this ?
$$\sum_{i=1} ^n (i^2+3i+1)\times i!= (n+3) \times (n+1)!-3$$
I did with induction, what I want to know is about other ways to prove this.
 A: I would use $$\sum_{i=1} ^n (i+2)! - i! = (n+2)!+(n+1)! - 2! - 1 !$$
A: As André Nicolas notes, the essence of the identity is showing that 
$$\sum_{k=1}^n k \cdot k! = (n+1)! - 1.$$
There's a nice combinatorial proof of this.  (See Benjamin and Quinn, Proofs that Really Count, Identity 181 on p. 92.)  I'll give it in its $\sum_{k=1}^{n-1} k \cdot k! = n! - 1$ form.  
Both sides count the number of permutations of $1, 2, \ldots, n$ that exclude the identity permutation.  
The right side is straightforward.  
For the left side, how many permutations have $n-k$ as the first number that does not get mapped to itself?  There are $k$ choices ($n-k+1, n-k+2, \ldots, n$) for the number that appears in position $n-k$, and then there are $k!$ ways to choose the remaining $k$ numbers to complete the permutation.  Adding up over all possible values of $k$ yields the left-hand side.
A: If you observe that $i^2 + 3i + 1 = (i + 2)(i+1) - 1$, then your left-hand sum is
$$\sum_{i=1}^n ((i+2)(i +1) - 1)i!$$
$$= \sum_{i=1}^n ((i + 2)! - i!)$$
As Henry noted, this sum telescopes into
$$(n + 2)! + (n + 1)! - 2! - 1!$$
$$= (n+2)(n+1)! + (n+1)! - 3$$
$$= (n+3)(n+1)! - 3$$
A: 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.  
