Question

If f(n+1)+f(n-1)=2f(n) and f(0)=0 then f(n) is (here n in N, the set of natural numbers)

Answer 1

Answer:

F(n) = 2 - 1/F(n-1) for all n greater than or equal to 2.

So, F(2) = 2 - 1/F(1) = 2 - 1/3 = 5/3

F(3) = 2 - 1/F(2) = 2 - 3/5 = 7/5

F(4) = 2 - 1/F(3) = 2 - 5/7 = 9/7 and so on.

Do you observe any pattern in the values? See if it matches with my observation:

F(n) = (2n+1)/(2n-1), for all natural numbers n.

(Plug in some values of n to verify!)

or, F(n) = 1 + 2/(2n-1)

Now,

F(1) + F(2) + F(3) + • • • +F(1000)

= 1000 + 2[ 1 + 1/3 + 1/5 + 1/7 + • • • + 1/1999]

= 1000 - 2∑(1/2n-1), as n runs from 1 to 1000.