= Solution
{c}
There are $n+2$ distinct <interpolation nodes> but only $n+1$ free <coefficients> in a <polynomial> of <polynomial degree> at most $n$. Thus <polynomial interpolation> is possible precisely when the data satisfy one compatibility equation. To identify it, form the <Lagrange interpolation polynomial> through all the data:
$$
q(x)=\sum_{i=0}^{n+1}f_i
\prod_{\substack{0\leq j\leq n+1\\j\ne i}}
\frac{x-x_j}{x_i-x_j}.
$$
This is the unique <interpolation polynomial> of <polynomial degree> at most $n+1$. Its <leading coefficient>, which is also the order-$(n+1)$ <divided difference>, is
$$
D=\sum_{i=0}^{n+1}\frac{f_i}
{\displaystyle\prod_{\substack{0\leq j\leq n+1\\j\ne i}}(x_i-x_j)}.
$$
Consequently the required condition is
$$
\boxed{D=0.}
$$
Necessity follows because any qualifying $p_n$ must equal $q$ by uniqueness of <polynomial interpolation>; sufficiency follows because $D=0$ removes the only possible term of <polynomial degree> $n+1$ from $q$.
When this condition holds, a slightly shorter expression uses only the first $n+1$ <interpolation nodes>:
$$
\boxed{p_n(x)=\sum_{i=0}^{n}f_i
\prod_{\substack{0\leq j\leq n\\j\ne i}}
\frac{x-x_j}{x_i-x_j}.}
$$
This <polynomial> agrees with $q$ at $n+1$ distinct points, so the <root bound for a polynomial> gives $p_n=q$ and hence $p_n(x_{n+1})=f_{n+1}$ as required. Equivalently, the compatibility condition says that this expression, evaluated at the last <interpolation node>, gives $f_{n+1}$.
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