Let $F$ be a function with domain $\mathbb Z^+$ such that $F(1)+F(2)+F(3)+......F(n)=n^2F(n)$. If $F(1)=2006$, then find the value of $F(2005)$

Question

A function F is defined for all the positive integers that satisfy the following condition:

$F(1)+F(2)+F(3)+......F(n)=n^2F(n)$. If $F(1)=2006$, then find the value of F(2005)

What I did:- putting n=2 we have $F(1)+F(2)=4*F(2)$ $F(2)=2006/3$

I did not understand from this step given in the solution

Putting n=3 we have $F(1)+F(2)+F(3)=9*F(3)$

$F(3)= 4/3*1/8=1/6 of 2006$

And then this step F(n)=1/(sum of all natural numbers till n)*2006 $F(2005)=2/(2005*2006)*2006$

I will be really grateful if anyone could explain me this solution in detail. I will be highly obliged

Answer 1

To understand the solution,try to prove by induction that $$1+2+...+n==\frac{n(n+1)}{2}$$

and take $n=2005$

Answer 2

Like you said:

$$F(1) + F(2) = 4F(2)\\ \implies 2006 +F(2) = 4F(2)\\ \implies 2006 = 3F(2)\\ \implies F(2) = \frac{2006}{3}$$

Going on to the next case:

$$F(1) + F(2) + F(3) = 9F(3)\\ \implies 2006 + \frac{2006}{3} + F(3) = 9F(3)\\ \implies \frac43\cdot 2006 = 8F(3)\\ \implies F(3) = \frac{2006}{6}$$

So, we have $F(1)=\frac{2006}{1}, F(2)=\frac{2006}{3}, F(3)=\frac{2006}{6}$. We'd like to identify the pattern there, and the denominators are doing something nice: they're $1, 1+2, 1+2+3$. You can verify this by computing $F(4)$, and seeing that it's $\frac{2006}{10}$.

Following this pattern, we expect $F(2005)$ to be $2006$ over $1+2+\cdots+2005$. Now the formula for $1+2+\cdots+k$ is $\frac{k(k+1)}{2}$. Dividing by a fraction is the same as multiplying by its reciprocal, so we get the given answer.