How to prove via induction.

Question

How would you prove (using induction) that:

(If $f(1) = 1996$)

$f(n) = \frac{1}{(2^{2} - 1)} \cdot \frac{2^2}{(3^{2} - 1)} \cdot \frac{3^2}{(4^2 - 1)} \cdot ... \cdot \frac{(n-1)^2}{n^{2} - 1} \cdot f(1)$ given that $f(1) + f(2) + f(3) + f(4) + ... + f(n)= n^2 f(n)$ and that $f$ is defined for all positive integers $n>1$?

Answer 1

Induction base holds, let's prove induction step.

Assume that holds $$f(n) = \frac{1}{(2^{2} - 1)} \cdot \frac{2^2}{(3^{2} - 1)} \cdot \frac{3^2}{(4^2 - 1)} \cdots \frac{(n-1)^2}{n^{2} - 1} \cdot f(1).$$ We have by our formula that $$(n+1)^2f(n+1)=f(1)+\dots+f(n)+f(n+1)=n^2f(n)+f(n+1),$$ from where we get $$((n+1)^2-1)f(n+1)=n^2f(n)$$ and $$f(n+1)=\frac{n^2}{(n+1)^2-1}f(n).$$ By using our assumption, we get $$f(n+1)=\frac{1}{(2^{2} - 1)} \cdot \frac{2^2}{(3^{2} - 1)} \cdot \frac{3^2}{(4^2 - 1)} \cdots\frac{(n-1)^2}{n^{2} - 1}\frac{(n+1-1)^2}{(n+1)^2-1} \cdot f(1).$$

Answer 2

Given that: $f(1) = 1996$ and $f(1) + f(2) + f(3) + f(4) + ... + f(n)= n^2 f(n)$

Prove by induction:

$f(n) = \frac{1}{(2^{2} - 1)} \cdot \frac{2^2}{(3^{2} - 1)} \cdot \frac{3^2}{(4^2 - 1)} \cdot ... \cdot \frac{(n-1)^2}{n^{2} - 1} \cdot f(1)$

base case $n=2$

$1996 + f(2) = 2^2 f(2)\\ (2^2 - 1) f(2) = 1996\\ f(2) = \frac 1{2^2+1} f(2)$

Inductive hypothesis:

Assume $f(n) = \frac{1}{(2^{2} - 1)} \cdot \frac{2^2}{(3^{2} - 1)} \cdot \frac{3^2}{(4^2 - 1)} \cdot ... \cdot \frac{(n-1)^2}{n^{2} - 1} \cdot f(1)$

We must show that

$f(n+1) = \frac{1}{(2^{2} - 1)} \cdot \frac{2^2}{(3^{2} - 1)} \cdot \frac{3^2}{(4^2 - 1)} \cdot ... \cdot \frac{(n-1)^2}{n^{2} - 1}\cdot \frac{(n)^2}{(n+1)^{2} - 1} \cdot f(1)$

Based on the inductive hypothesis.

$f(1) + f(2) + f(3) + f(4) + ... + f(n) + f(n+1)= (n+1)^2 f(n+1)\\ f(1) + f(2) + f(3) + f(4) + ... + f(n)= ((n+1)^2 -1) f(n+1)\\ n^2 f(n)= ((n+1)^2 -1) f(n+1)\\ n^2\frac{1}{(2^{2} - 1)} \cdot \frac{2^2}{(3^{2} - 1)} \cdot \frac{3^2}{(4^2 - 1)} \cdot ... \cdot \frac{(n-1)^2}{n^{2} - 1} \cdot f(1)=((n+1)^2 -1) f(n+1)\\ f(n+1) = \frac{(n)^2}{(n+1)^{2} - 1} \cdot\frac{1}{(2^{2} - 1)} \cdot \frac{2^2}{(3^{2} - 1)} \cdot \frac{3^2}{(4^2 - 1)} \cdot ... \cdot \frac{(n-1)^2}{n^{2} - 1} \cdot f(1)$

QED

Answer 3

Suppose it is true for $n \in \Bbb N$. Then $$n^2f(n)+f(n+1)=f(1)+...+f(n)+f(n+1)=(n+1)^2f(n+1) \iff f(n+1) = \frac{n^2}{(n+1)^2-1}f(n)$$