f(n)=1!+2!+......n! n belongs to natural numbers.

Find P(n) & Q(n) such that f(n+2)= P(n)f(n+1)+Q(n)f(n)

f(n)=1!+2!+......n!  n belongs to natural numbers.

f(n+2)= P(n)f(n+1)+Q(n)f(n)

1!+2!+......+n!+(n+1)!+(n+2)! = P(n)*[1!+2!+......+n!+(n+1)!] + Q(n)*[1!+2!+......+n!]

                                           = [1!+2!+......+n!]*  + P(n)*(n+1)!

on comparing both sides, we have [P(n)+Q(n)] = 1

         and P(n)*(n+1)! = (n+1)!+(n+2)! = (n+1)! * (n+3)

            P(n)=n+3

           Q(n)=-n-2

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Regards

Ramesh

f(n+2) = f(n) + (n+1)! + (n+2)! = f(n) + (n+1)! + (n+2) (n+1)! = f(n) + (n+3)(n+1)!...............1

f(n+2) = f(n+1) + (n+2)! = f(n+1) + (n+2)(n+1)!..........................2

Hence (n+3) [f(n+2) - f(n+1)] = (n+2) [f(n+2) - f(n)] which yields

f(n+2) = (n+3) f(n+1) - (n+2) f(n)