# For n>1, f(1)=2, f(1)+f(2)+f(3)+f(4)+...+f(n)=n^2f(n) Find the exact value of f(1000)? and I think that f(2)=2/3 and f(3)=1/3 (can you please show steps and formula's used, from where you got...

For n>1, f(1)=2, f(1)+f(2)+f(3)+f(4)+...+f(n)=n^2f(n)

**Find the exact value of f(1000)?**

and I think that f(2)=2/3 and f(3)=1/3

(can you please show steps and formula's used, from where you got the answer? It would really help my understanding on this tough question)

Thanks for answering/reading my question. :)

### 1 Answer | Add Yours

This is a tough question. I do not see an immediate use of a nice formula to get the answer, but I can show you some of the steps I took:

(1) Find the first few values of the function. We hope to find some pattern, or better a closed form analytic function:

f(1)=2 is given.

For f(2) we have f(1)+f(2)=4f(2) or 2=3f(2) ==> f(2)=2/3.

f(3): 2+2/3+f(3)=9f(3) ==> 8/3=8f(3) ==> f(3)=1/3.

f(4): 2+2/3+1/3+f(4)=16f(4) ==> 3=15f(4)==> f(4)=1/5

f(5):16/5+f(5)=24f(5)==> f(5)=2/15

f(6): 10/3+f(6)=35f(6)==> f(6)=2/21

f(7): 24/7+f(7)=48f(7) ==> f(7)=1/14

So we have the sequence 2,2/3,1/3,1/5,2/15,2/21,1/14,... as the values of f(n)

(2) I tried a lot of different things before I hit on the answer. You might consider trying to write f(n) in terms of f(n-1) -- it turns out to be relatively simple but unuseful as we have to know f(999) in order to compute f(1000); this is the problem with recursive definitions.

(3) Look at the sequence again -- this time in terms of the first term 2:

2=2

2/3=2*1/3

1/3=2*1/6

1/5=2*1/10

2/15=2*1/15

2/21=2*1/21

1/14=2*1/28

Do you see the pattern? Consider the denominators.

1,3,6,10,15,21,28,... is the first few terms of the triangular numbers. The triangular numbers are formed by:

1

1+2

1+2+3

1+2+3+4

etc... This looks like the formulation for our function. I haven't tried to prove the connection, but it works for the next several terms so it is likely that it works.

The nice thing is that there is a closed form formula for the nth triangular number -- `T(n)=(n(n+1))/2` **See link**

(4) So `f(1000)=2*1/((1000*1001)/2)=1/250250`

**Thus the exact value for f(1000) is `1/250250` **

**Sources:**