Dancing Triplets

$a_{n+1}a_{n+2}-2a_{n}a_{n+2}+a_{n}a_{n+1}=\cfrac{2a_{n}a_{n+1}a_{n+2}}{n(n+1)(n+2)}$

Do note that $a_{n}\neq0$ because if $a_{n}=0$ , $a_{n+1}=0$ or $a_{n+2}=0$ , which could never hold concurrently with $a_{1}\neq0$ and $a_{2}\neq0$ .

$\Rightarrow \cfrac{a_{n+1}a_{n+2}-2a_{n}a_{n+2}+a_{n}a_{n+1}}{a_{n}a_{n+1}a_{n+2}}=\cfrac{2}{n(n+1)(n+2)}$

$\cfrac{1}{a_{n}}-\cfrac{2}{a_{n+1}}+\cfrac{1}{a_{n+2}}=\cfrac{2}{n(n+1)(n+2)}$

By Partial Fraction Decomposition ,

$\cfrac{2}{n(n+1)(n+2)}=\cfrac{1}{n}-\cfrac{2}{n+1}+\cfrac{1}{n+2}$

$\cfrac{1}{a_{n}}-\cfrac{2}{a_{n+1}}+\cfrac{1}{a_{n+2}}=\cfrac{1}{n}-\cfrac{2}{n+1}+\cfrac{1}{n+2}$

We deduce that $a_{n}=n$

Indeed,

$a_{1}=1,a_{2}=2$ ,

which is sufficient to prove that the sequence holds.

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$f(2018)$

$=\sum_{k=1}^{2018} (-1)^{k}\cdot (\prod_{i=1}^{k}(a_{i}a_{i+1})^{(-1)^{i}})^{(-1)^{k}}$

$=(-1)^{1}\cdot (\prod_{i=1}^{1}(a_{i}a_{i+1})^{(-1)^{i}})^{(-1)^{1}}$

$+(-1)^{2}\cdot (\prod_{i=1}^{2}(a_{i}a_{i+1})^{(-1)^{i}})^{(-1)^{2}}+...$

$+(-1)^{2017}\cdot (\prod_{i=1}^{2017}(a_{i}a_{i+1})^{(-1)^{i}})^{(-1)^{2017}}$

$+(-1)^{2018}\cdot (\prod_{i=1}^{2018}(a_{i}a_{i+1})^{(-1)^{i}})^{(-1)^{2018}}$

$=- \cfrac{1}{\prod_{i=1}^{1}(a_{i}a_{i+1})^{(-1)^{i}}}+\prod_{i=1}^{2}(a_{i}a_{i+1})^{(-1)^{i}}+...- \cfrac{1}{\prod_{i=1}^{2017}(a_{i}a_{i+1})^{(-1)^{i}}}+\prod_{i=1}^{2018}(a_{i}a_{i+1})^{(-1)^{i}}$

$=- \cfrac{1}{\prod_{i=1}^{1}(i(i+1))^{(-1)^{i}}}+\prod_{i=1}^{2}(i(i+1))^{(-1)^{i}}+...- \cfrac{1}{\prod_{i=1}^{2017}(i(i+1))^{(-1)^{i}}}+\prod_{i=1}^{2018}(i(i+1))^{(-1)^{i}}$

$=- \cfrac{1}{\cfrac{1}{1\cdot2}}+\cfrac{1}{1\cdot2}\cdot2\cdot3+...-\cfrac{1}{\cfrac{1}{1\cdot2}\cdot2\cdot3+...\cdot\cfrac{1}{2017\cdot2018}}+\cfrac{1}{1\cdot 2}\cdot2\cdot3\cdot...\cdot\cfrac{1}{2017\cdot2018}\cdot2018\cdot2019$

$=- \cfrac{1}{\cfrac{1}{1\cdot2}}+\cfrac{1}{1\cdot2}\cdot2\cdot3-\cfrac{1}{\cfrac{1}{1\cdot2}\cdot2\cdot3\cdot\cfrac{1}{3\cdot4}}+...-\cfrac{1}{\cfrac{1}{1\cdot2}\cdot2\cdot3\cdot\cfrac{1}{3\cdot4}+...\cdot2016\cdot2017\cdot\cfrac{1}{2017\cdot2018}}+\cfrac{1}{1\cdot 2}\cdot2\cdot3\cdot\cfrac{1}{3\cdot4}\cdot...\cdot\cfrac{1}{2017\cdot2018}\cdot2018\cdot2019$

$=-\cfrac{1}{\cfrac{1}{2}}+3-\cfrac{1}{\cfrac{1}{4}}...-\cfrac{1}{\cfrac{1}{2018}}+2019$

$=(-2+3)+(-4+5)+...+(-2018+2019)$

$=1\cdot\cfrac{2018}{2}$

$=1009$