A Question on Polynomials

Algebra Level 4

Given that $f$ is a polynomial of degree 100 and that $f(k)=k$ for $k=1, 2, \ldots , 100$ and $f(101)=102$ . Find $f(102)$ .

1 solution

Chan Lye Lee
Mar 31, 2016

Let $\displaystyle f(x)=x+c\prod_{i=1}^{100}(x-i)$ where $c$ is a constant. Then it fulfill the condition that $f(k)=k$ for $k=1, 2, \ldots , 100$ .

Since $f(101)=102$ , it means that $102=101+c\times 100!$ and hence $c=\frac{1}{100!}$ .

Now $\displaystyle f(x)=x+\frac{1}{100!}\prod_{i=1}^{100}(x-i)$ and thus $f(102)=102+\frac{1}{100!}\times (101!)=203$ .