Let n be a positive integer with f(n)=1!...

Let n be a positive integer with `f(n)=1!+2!+3!+….+n!` and P(x) and Q(x) be polynomials in x such that `f(n+2) = Q(n)f(n) + P(n)f(n+1)`. Find P(2).

P(x)=x+3

Q(x)=-1+2

P(X)=-1-2

Q(x)=x+3

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