Question: > 
> Let n be a positive integer with $f(n)= 1! +2! +3!+... +n!$ and P(x),
>  Q(x) be polynomials in $x$ such that $f(n+2)=P(n)f(n+1)+Q(n)f(n)$ for
>  all $n \geq 1$, then which of the options **is/are** correct? 
> 
> 
> 1. $P(x)= x+3$
> 2. $Q(x)= -x-2$
> 3. $P(x)= -x-2$
> 4. $Q(x)= x+3$
> 
> 
> 

I managed to get that options 1 and 2 are correct by calculating $f(1),f(2),f(3),f(4)$, substituting them in the equation and then trying out all the options. Is there any way to do the problem without using the options?

Answer: 
$$
\big[1!+2!+\dots+n!+(n+1)!+(n+2)!\big]=\\
\big[1!+2!+\dots+n!+(n+1)!\big]P(n)+\\
\big[1!+2!+\dots+n!\big]Q(n)
$$
So an easy way to get this is to have $P(n)+Q(n)=1$ so that all terms
$1!+2!+\dots+n!$ come out right. Then we want $P(n)$ so that
$$
(n+1)!P(n) = (n+1)!+(n+2)!
$$
Divide by $(n+1)!$ to get
$$
P(n) = 1+(n+2)=n+3.
$$
and, as noted $P(n)+Q(n)=1$, so $Q(n) = -n-2$.
If $P(x)$ is greater than linear or $Q(x)$ is greater than quadratic you have a delicate cancellation which can not be achieved. You have to have $Q$ one degree higher than $P$. Now look at lots of points for large $n$ and argue the difference cannot have that many roots. I am handwaving, but I believe it. Given that, we have five constants to determine, so can just take five values and solve the equations 
$$10=3P(2)+Q(1)\\34=10P(3)+3Q(2)$$ and so on. Now let $P(n)=an+b, Q(n)=cn^2+dn+e$ and you get five equations in five unknowns. 

In this case, the high order term comes only from $P(n)$, which forces that $P(n)=n+b$ and $Q(n)=dn+3$ so you only need three equations. I also don't know how to prove that.