# If $f(x)$ denotes a polynomial of degree $n$ such that $f(k) =\dfrac1k$ for $k = 1,2,3,\ldots,(n+1)$, determine $f(n + 2)$.

Let $f(x)$ be a polynomial of degree $n$ such that $$f(k) =\dfrac1k$$ for $k = 1,2,3,\ldots,(n+1)$. Determine $f(n + 2)$.

This question is quite similar to this one. How do I solve such type of problems in general?

• $P(k)=k/(k+1)$ for an interesting range of numbers if and only if $1-P(k)=1/(k+1)$ for that same range. Looks like you may be looking for $f(x)=1-P(x-1)$ where $P(x)$ the polynomial that popped out in the linked question. Adjust $n$ by one. – Jyrki Lahtonen Jul 12 '17 at 10:58

Hint: Note that we have $kf(k) = 1$ for $k = 1,\ldots,n+1$, and clearly $0f(0) = 0$. So now we know $n+2$ values of the $(n+1)$-degree polynomial $xf(x)$.
As for the general bit, it's difficult to say. You should try to make everything into polynomials as nicely as possible, because polynomials are a lot easier to work with than rational functions. Case in point: I can't tell much about $f(x)$ just from knowing the solutions of $f(x) = \frac1x$, but I can tell a lot about $xf(x)$ from knowing the solutions to $xf(x) = 1$.
Let $g(x):=x\,f(x)$ like Arthur suggests. Then, $g$ is of degree $n+1$. Hence, from this link, we have $$\sum_{r=0}^{n+2}\,(-1)^r\,\binom{n+2}{r}\,g(x+r)\equiv 0\,.$$ In particular, $$\sum_{r=0}^{n+2}\,(-1)^r\,\binom{n+2}{r}\,g(r)=0\,.$$ As $g(0)=0$ and $g(1)=g(2)=\ldots=g(n+1)=1$, we conclude that \begin{align}g(n+2)&=(-1)^{n+1}\,\sum_{r=1}^{n+1}\,(-1)^{r}\,\binom{n+2}{r} \\&=(-1)^{n+1}\,\left((1-1)^{n+2}-1-(-1)^{n+2}\right)=(-1)^{n}+1\,.\end{align} Consequently, $$f(n+2)=\frac{1+(-1)^n}{n+2}\,.$$