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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 -- Regards Ramesh
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
--
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)
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)
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