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harshit agarwal Grade: 12
        

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)

8 years ago

Answers : (2)

Ramesh V
70 Points
										

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

8 years ago
mycroft holmes
266 Points
										

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)

8 years ago
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