If $1(0!)+3.(1!)+7(2!)+13(3!) +21(4!) + \cdots $ n terms.... Question( from sequences) : 
If $1(0!)+3.(1!)+7(2!)+13(3!) +21(4!) + \cdots $ n terms = $(4000)(4000!)$ Then what is the value of n. 
How to proceed in this please suggest , will be of great help to me thanks...
 A: It is a telescoping sum.
$$\begin{align}
\sum_{k=0}^n k!(k(k+1)+1)
=& \sum_{k=0}^n k!\big((k+1)(k+2) - 2(k+1) + 1\big)\\
=& \sum_{k=0}^n \big((k+2)! -2(k+1)! + k!\big)\\
=& \sum_{k=0}^n \bigg(\big((k+2)! - (k+1)!\big) - \big((k+1)!-k!\big)\bigg)\\
=& \big((n+2)! - (0+1)!\big) - \big((n+1)! - 0!\big)\\
=& (n+2)! - (n+1)!\\
=& (n+1)(n+1)!\\
\end{align}
$$
So $n = 3999$.
A: The summation is
\begin{align}
S_{n} = \sum_{r=0}^{n} \left( r(r+1) + 1\right) \, r!
\end{align}
and by selecting $n$ values it is seen that
\begin{align}
S_{0} &= 1 \\
S_{1} &= 0! + 3 \cdot 1! = 2 \cdot 2! \\
S_{2} &= 0! + 3 \cdot 1! + 7 \cdot 2! = 3 \cdot 3!
\end{align}
which leads to
\begin{align}
S_{n} = \sum_{r=0}^{n} [ r(r+1) + 1] \, r! = (n+1) \cdot (n+1)!.
\end{align}
For the case of $S_{n} = (4000) \cdot (4000)!$ it is seen that $n = 3999$. 
A: Suggestions: I would look for patterns:
$1(0!)=1$
$1(0!)+3(1!)=4$
$1(0!)+3(1!)+7(2!)=18$.
Then I'd try to match this to something that would give $(4000)(4000!)$ for some $n$.  Then I'd start trying to prove some stuff (probably by induction).
A: Your approach should be to write a generic formula with a summation.
Then figure out that each summation element adds a new element but removes the old one.
And if you do the administration properly, you'll find n to be 3999 or 4000 or so.
