Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 12F / iii / a / Solution

The coefficient of $x^{n+1}$ in $r_j$ isWith the increasing enumeration, its sign is $(-1{)}^{n+2-j}$. Therefore all terms $(-1{)}^j{c}_j$ have the same sign $(-1{)}^{n+2}$, and $R:={\sum }_j(-1{)}^j{c}_j\ne 0$. The leading coefficient of $t$ is $T:={\sum }_{j}f(y_j)c_j$. Canceling the coefficient of $x^{n+1}$ gives the unique valueBoth interpolants have degree at most $n+1$, so this single cancellation is equivalent to $\mathrm{deg}(t-\lambda r)\le n$.