<s>[INST] 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...
 [/INST] 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 $n = 3999$. 
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