Happy birthday Number theory!

\[\sum _{ k=1 }^{ n }{ \left( \frac { 1 }{ f(k)f(k+1) } \right) } =\frac { f(f(n)) }{ f(n+1) } \]

A function \(f : \mathbb {N \to N} \) satisfies above conditions \(\forall n \in \mathbb N \).

Find \(f(2017)\).

×

Problem Loading...

Note Loading...

Set Loading...