[Guest]

F is a function such that:

F(1) + 2*F(2) + 3*F(3) + .......+ N*F(N) = N*(N+1)*F(N), whenvever N is a positive integer >=2, and:

F(1) = 1

Determine the value of F(2005).

[Guest]

Here's my thought process, tell me where I'm messing up:

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) + N*F(N) = N*F(N)*(N+1)

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) = N*F(N)*(N+1) - N*F(N)

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) = N*F(N)*(N+1-1)

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) = N²*F(N)

(N-1)(N+1-1)F(N-1) = N²*F(N)

F(N) = (N-1)/(N) * F(N-1)

F(N) = (N-1)!/(N)! * F(1)

F(N) = 1/N * F(1) = 1/N

But when I plug this into the original equation it's always off by 1...

[Guest]

WHEN N=2,

F(1)+2*F(2)=6*F(2)

IMPLIES,1=4*F(2)

IMPLIES,F(2)=1/4,

SIMILARLY,

WHEN N=3,

F(1)+2*F(2)+3*F(3)=12*F(3),

1+1/2+3*F(3)=12*F(3),

3/2=9*F(3),

F(3)=1/6

SO, WHEN N=2005,

F(2005)=1/(2*2005)=1/4010

[Guest]

...by induction.

So we have F(1) = 1. Plugging N = 2 into the formula and then solving for F(2) gives us F(2) = 1/4, and similarly, F(3) = 1/6. We will guess that in general, for all integers N > 1, F(N) = 1/(2N).

Proof:

The closed formula works for N = 2 and N = 3. Suppose that for all integers in {1, 2, ..., K), that F(K) = 1/(2K). Then, note that K*F(K) = K/(2K) = 1/2. Thus, we have:

F(1) + 2*F(2) + ... + K*F(K) + (K+1)*F(K+1) = (K+1)*(K+2)*F(K+1)

1 + 1/2 + 1/2 + ... + (K+1)*F(K+1) = (K+1)*(K+2)*F(K+1)

There are K+1 terms on the left side, all but two of which are 1/2, so there are K-1 terms that are 1/2...

1 + (K-1)*(1/2) + (K+1)*F(K+1) = (K+1)*(K+2)*F(K+1)

1 + (K-1)/2 + (K+1)*F(K+1) = (K+1)*(K+2)*F(K+1)

(K+1)/2 + (K+1)*F(K+1) = (K+1)*(K+2)*F(K+1)

1/2 + F(K+1) = (K+2)*F(K+1)

1/2 = (K+1)*F(K+1)

1/(2*(K+1)) = F(K+1)

So if for all N in {1,2,...,K}, F(N) = 1/(2N), then F(K+1) = 1/(2*(K+1)). By induction, for all integers N > 1, F(N) = 1/(2N), and F(2005) = 1/4010.

[superprismatic]

Here's my thought process, tell me where I'm messing up:

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) + N*F(N) = N*F(N)*(N+1)

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) = N*F(N)*(N+1) - N*F(N)

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) = N*F(N)*(N+1-1)

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) = N²*F(N)

(N-1)(N+1-1)F(N-1) = N²*F(N)

F(N) = (N-1)/(N) * F(N-1)

F(N) = (N-1)!/(N)! * F(1)

F(N) = 1/N * F(1) = 1/N

But when I plug this into the original equation it's always off by 1...

I think the problem lies in the fact that F(1) doesn't follow

the general rule because F(1) does not equal 1*F(1)*(1+1).

So stop at F(2) like so (modifying your stuff):

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) + N*F(N) = N*F(N)*(N+1)

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) = N*F(N)*(N+1) - N*F(N)

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) = N*F(N)*(N+1-1)

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) = N2*F(N)

(N-1)(N+1-1)F(N-1) = N2*F(N)

F(N) = (N-1)/(N) * F(N-1)

F(N) = [(N-1)/N] * [(N-2)/(N-1)] * ... * [3/4] * [2/3] * F(2)

F(N)=2*F(2)/N

and since direct computation gives F(2)=1/4, you have

F(N)=1/(2*N)

I like the way you proceeded. Very Nice!

[Guest]

I think the problem lies in the fact that F(1) doesn't follow

the general rule because F(1) does not equal 1*F(1)*(1+1).

So stop at F(2) like so (modifying your stuff):

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) + N*F(N) = N*F(N)*(N+1)

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) = N*F(N)*(N+1) - N*F(N)

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) = N*F(N)*(N+1-1)

F(1) + 2F(2) + 3F(3) + ... + (N-1)F(N-1) = N2*F(N)

(N-1)(N+1-1)F(N-1) = N2*F(N)

F(N) = (N-1)/(N) * F(N-1)

F(N) = [(N-1)/N] * [(N-2)/(N-1)] * ... * [3/4] * [2/3] * F(2)

F(N)=2*F(2)/N

and since direct computation gives F(2)=1/4, you have

F(N)=1/(2*N)

I like the way you proceeded. Very Nice!

Ah yes. That does make sense. Many thanks!