Difficult Polynomial Question

Question

Let $P(x)$ be a polynomial whose degree is 1996. If $P(n) = \frac{1}{n}$ for $n = 1, 2, 3, . . . , 1997$, compute the value of $P(1998).$

I don't even know where to begin...

Any and all help would be appreciated, thanks!

Answer 1

If $P(n) = \frac 1n$ for $n=1,2,3,\dots,1997$, then $nP(n)-1=0$ for all $1 \le n \le 1997$. That is, the $1997^{th}$ degree polynomial $xP(x)-1$ has roots $1,2,3,\dots,1997$.

This implies $xP(x)-1=c(x-1)(x-2)\dots(x-1997)$. If $x=0$, we have $-1=-c(1997!)$.

Hence,

$$c = \frac 1{1997!}\implies 1998P(1998)-1 = \frac 1{1997!}(1997!)=1 \implies P(1998)=\frac 2{1998}$$

(This is more or less verbatim from Chantry Cargill's link above. I wrote it as a CW answer to avoid linkrot).