TITLE: For any natural $n$, how could we prove that $\sum\limits_{i=1} ^n (i^2+3i+1) i!= (n+3)(n+1)! - 3$
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: 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.

REPLY [3 votes]: 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
$$(n + 2)! + (n + 1)! - 2! - 1!$$
$$= (n+2)(n+1)! + (n+1)! - 3$$
$$= (n+3)(n+1)! - 3$$