6. It is given that the sequence $\left(a_{n}\right)_{n=1}^{\infty}$, with $a_{1}=a_{2}=2$, is given by the recurrence relation
$$
\frac{2 a_{n-1} a_{n}}{a_{n-1} a_{n+1}-a_{n}^{2}}=n^{3}-n
$$
for all $n=2,3,4, \ldots$. Find the integer that is closest to the value of $\sum_{k=2}^{2011} \frac{a_{k+1}}{a_{k}}$.
6. Answer: 3015
Solution. The recurrence relation can be written as
$$
\frac{a_{n+1}}{a_{n}}-\frac{a_{n}}{a_{n-1}}=\left(\frac{1}{n-1}-\frac{1}{n}\right)-\left(\frac{1}{n}-\frac{1}{n+1}\right) .
$$

Summing for $n=2$ to $N$, we obtain
$$
\frac{a_{N+1}}{a_{N}}-\frac{a_{2}}{a_{1}}=\left(1-\frac{1}{N}\right)-\left(\frac{1}{2}-\frac{1}{N+1}\right),
$$
showing that
$$
\frac{a_{N+1}}{a_{N}}=\frac{3}{2}-\left(\frac{1}{N}-\frac{1}{N+1}\right) .
$$

Summing this up for $N=2$ to $N=2011$, we obtain
$$
S=\sum_{k=2}^{2011} \frac{a_{k+1}}{a_{k}}=\frac{3}{2}(2010)-\left(\frac{1}{2}-\frac{1}{2012}\right)=3014.5+\frac{1}{2012}
$$
showing that the integer closest to $S$ is 3015 .