f(n)=( 1/√5)*{[( 1+√5)/2]^(n- 1)[( 1-√5)/2]^(n- 1)}

Let n= 100 1,

f( 100 1)= 0

( 1/√5)*{[( 1+√5)/2]^( 1000) - [( 1-√5)/2]^( 1000)}

The following is the formula solution (excerpt):

Let constant r, s

Let f (n)-r * f (n-1) = s * [f (n-1)-r * f (n-2)]

Then r+s= 1, -rs= 1.

When n≥3, there are

F(n)-r * F(n- 1)= s *[F(n- 1)-r * F(n-2)]

F(n- 1)-r * F(n-2)= s *[F(n-2)-r * F(n-3)]

F(n-2)-r * F(n-3)= s *[F(n-3)-r * F(n-4)]

……

F(3)-r * F(2)= s *[F(2)-r * F( 1)]

Multiply the above n-2 expressions to get:

f(n)-r*f(n- 1)=[s^(n-2)]*[f(2)-r*f( 1)]

∫s = 1-r，F( 1)=F(2)= 1

The above formula can be simplified as:

f(n)=s^(n- 1)+r*f(n- 1)

So:

f(n)=s^(n- 1)+r*f(n- 1)

= s^(n- 1)+r*s^(n-2)+r^2*f(n-2)

= s^(n- 1)+r*s^(n-2)+r^2*s^(n-3)+r^3*f(n-3)

……

= s^(n- 1)+r*s^(n-2)+r^2*s^(n-3)+……+r^(n-2)*s+r^(n- 1)*f( 1)

= s^(n- 1)+r*s^(n-2)+r^2*s^(n-3)+……+r^(n-2)*s+r^(n- 1)

(This is the sum of terms of geometric series with S (n- 1) as the first term, R (n- 1) as the last term, and r/s as the tolerance).

=[s^(n- 1)-r^(n- 1)*r/s]/( 1-r/s)

=(s^n - r^n)/(s-r)

The solutions of R+S = 1 and -RS = 1 are S = (1+√ 5)/2, and R = (1-√ 5)/2.

Then f (n) = (1√ 5) * {[(1+√ 5)/2] n-[(1-√ 5)/2] n}

2.An=2+4(n- 1)

When n= 10004

An=400 14