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Question

Let n be a positive integer and define f(n)=1!+2!+3!++n!.Find Polynomials P(x) and Q(x) such that f(n+2)=Q(n)f(n)+P(n)f(n+1) forall n1
Find P(2).

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Solution

f(n)=1!+2!+3!++n!
f(n+2)=Q(n)f(n)+P(n)f(n+1)
1!+2!++(n+2)!=Q(n)[1!+2!++n!]+P(n)[1!+2!++(n+1)!]
Comparing coefficients upto (n1)!
Q(n)+P(n)=1Q(n)=1P(n)
n!+(n+1)!+(n+2)!=(n!)Q(n)+(n!+(n+1)!)P(n)
1+n+1+n2+3n+2=1P(n)+(n+2)P(n)
P(n)=n+3Q(n)=(n+2)
P(2)=5

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