Finding $f(100)$ given that $f$ is a polynomial of degree $98$ and $f (k)=\frac 1 k$ for $k=1,2,3,\dots,98,99$

Question

A polynomial of degree $98$ such $f (k)=\frac 1 k$ for $k=1,2,3,\dots,98,99$ exists. How can I find $f(100)$? What are the possible methods?

Answer 1

This is based on Akiva Weinberger columbus's hint.

Consider the polynomial $$kf(k)-1$$

which has degree 99 is zero for all $k=1,2,3, \cdots, 99$.

So you know the polynomial $kf(k)-1$ has the form $$C(k-1)(k-2)(k-3)\cdots(k-99)$$

and you can find $C$ when you put $k=0$.

In general, its a good idea to find a polynomial that has as many zeros as its degree is.