a function f is defined for all positive integers and satisfies f(1)=2005 and f(1)+f(2)—-——f(n)=n^2f(n) for all n>1.find value of f(2004).please explain the methodology.
YES,NOW DO TELL ME THYE METHODOLOGY
IT FORMS A G.P. THEN WE SOLVE IT.
NOT A G.P. BUT, A SEQUENCE IS FORMED.
YEAH,SOMEWHAT BUT STILL BEATING AROUND THE BUSH
By the given definition , we have
From (1) and (2), we have
Now, we have
Sorry friend the above solution is wrong
The correct solution is here
given that f(1)=2005
and for n > 1 f(1) + f(2) + f(3) + ....... +f(n) + = f(n) .......... (1)
Equation 1 can be written as
Now =
f(n-1).... (3)
( Since from equation 1)Now substitute the vale of equation 3 in equation 2
We get f(n-1) +f(n)=
f(n)
This can be written as (n-1)(n-1) .f(n-1) = .f(n) – f(n)
(n-1)(n-1) .f(n-1) = .f(n)
(n-1)(n-1) .f(n-1) = .f(n)
(n-1).f(n-1) = (n+1).f(n)
now given that f(1)=2005
Now to find f(2) put n =2 in equation (4)
We get =
f(2)= =
( since f(1) =2005 )
Now to find f(3) put n =3 in equation (4)
We get =
f(3)= =
.
( since f(2) =
)
Now to find f(4) put n =4 in equation (4)
We get =
f(4)= =
.
.
( since f(3)==
.
Now to find f(5) put n =5 in equation (4)
We get =
f(5)= =
.
.
.
( since f(4)=
.
.\frac{1.2005}{3}
f(5) = .
.
.
This is is true for first 5 terms
The above equation can be written as
f(5) = .
.
.
If we generalise for n terms
The numarator is 1.2.3.4…....(n-1) = (n-1)!
And the denominator is 1.2.3.4.5.6…..( n+1) = (n+1)! ( by 0bserving the above statement )
so f(n) = =
f(n) =
f(n) =
Now substitute n = 2004 in the above equation
We have f(2004) =
f(2004) = 
f(2004) = 
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