Year: 2009
Paper: 3
Question Number: 5
Course: UFM Pure
Section: Roots of polynomials
The vast majority of candidates (in excess of 95%) attempted at least five questions, and nearly a quarter attempted more than six questions, though very few doing so achieved high scores (about 2%). Most attempting more than six questions were submitting fragmentary answers, which, as the rubric informed candidates, earned little credit.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
The numbers $x$, $y$ and $z$ satisfy
\begin{align*}
x+y+z&= 1\\
x^2+y^2+z^2&=2\\
x^3+y^3+z^3&=3\,.
\end{align*}
Show that
\[
yz+zx+xy=-\frac12 \,.\]
Show also that $x^2y+x^2z+y^2z+y^2x+z^2x+z^2y=-1\,$,
and hence that
\[
xyz=\frac16 \,.\]
Let $S_n=x^n+y^n+z^n\,$.
Use the above results to
find numbers $a$, $b$ and $c$
such that the relation
\[
S_{n+1}=aS_{n}+bS_{n-1}+cS_{n-2}\,,
\]
holds for all $n$.\begin{align*}
&& (x+y+z)^2 &= x^2 + y^2 + z^2 + 2(xy+yz+zx) \\
\Rightarrow && 1^2 &= 2 + 2(xy+yz+zx) \\
\Rightarrow && xy+yz+zx &= -\frac12
\end{align*}
\begin{align*}
&& 1 \cdot 2 &= (x+y+z)(x^2+y^2+z^2) \\
&&&= x^3 + y^3 + z^3 + x^2y+x^2z+y^2z+y^2x+z^2x+z^2y \\
&&&= 3 + x^2y+x^2z+y^2z+y^2x+z^2x+z^2y\\
\Rightarrow && -1 &= x^2y+x^2z+y^2z+y^2x+z^2x+z^2y
\end{align*}
\begin{align*}
&& (x+y+z)^3 &= x^3 + y^3 + z^3 + \\
&&&\quad \quad 3xy^2 + 3xz^2 + \cdots + 3zx^2 + 3zy^2 + \\
&&&\quad \quad \quad 6xyz \\
\Rightarrow && 1 &= 3 + 3(-1) + 6xyz \\
\Rightarrow && xyz &= \frac16
\end{align*}
Since we have $f(t) = (t-x)(t-y)(t-z) = t^3-t^2-\frac12 t - \frac16$ is zero for $x,y,z$ we can notice that:
$t^{n+1} = t^n +\frac12 t^{n-1} + \frac16 t^{n-2}$ is also true for $x,y,z$ (by multiplying by $t^{n-2}$.
Therefore:
$S_{n+1} = S_n + \frac12 S_{n-1} + \frac16 S_{n-2}$
This was the most popular question, with a few more attempts than question 3, and with a level of success matching the first two questions. Many showed the first two results correctly, and quite a few the third one. The last part tripped up many candidates, though the most successful used the first approach in the mark scheme. A number of candidates understood "independent of n" in the question to be given information, and attempted to find a, b, and c by solving three simultaneous equations for specific values of n. However, there were commonly errors in the values of the Sₙ used. An efficient alternative solution is given in the mark scheme.