1.
Let \(a_0=-2,b_0=1\), and for\( n\ge 0, let \begin{align*}a_{n+1}&=a_n+b_n+\sqrt{a_n^2+b_n^2},\\ b_{n+1}&=a_n+b_n-\sqrt{a_n^2+b_n^2}.\end{align*}\)
Find\( \dfrac{1}{a_{2012}} + \dfrac{1}{b_{2012}}\).
\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{a_{n+1}} + \dfrac{1}{b_{n+1}}} &=& \dfrac{1}{a_n+b_n+\sqrt{a_n^2+b_n^2}} + \dfrac{1}{a_n+b_n-\sqrt{a_n^2+b_n^2}} \\\\ &=& \dfrac{a_n+b_n-\sqrt{a_n^2+b_n^2}+a_n+b_n-\sqrt{a_n^2+b_n^2}} {\left(a_n+b_n+\sqrt{a_n^2+b_n^2}\right)\left(a_n+b_n-\sqrt{a_n^2+b_n^2}\right)} \\\\ &=& \dfrac{2(a_n+b_n)} {\left(a_n+b_n\right)^2-\left(\sqrt{a_n^2+b_n^2}\right)^2} \\\\ &=& \dfrac{2(a_n+b_n)} {a_n^2+2a_nb_n+b_n^2-(a_n^2+b_n^2)} \\\\ &=& \dfrac{2(a_n+b_n)} { 2a_nb_n} \\\\ &=& \dfrac{ a_n+b_n } { a_nb_n} \\\\ &=& \mathbf{\dfrac{1}{a_{n}} + \dfrac{1}{b_{n}}} \\ \hline \mathbf{\dfrac{1}{a_{n+1}} + \dfrac{1}{b_{n+1}}} &=& \mathbf{\dfrac{1}{a_{n}} + \dfrac{1}{b_{n}}} = \ldots = \mathbf{\dfrac{1}{a_{1}} + \dfrac{1}{b_{1}}} \\\\ \dfrac{1}{a_{1}} + \dfrac{1}{b_{1}} &=& \dfrac{1}{-1+\sqrt{5}} + \dfrac{1}{-1-\sqrt{5}} = \dfrac{1}{2} \\\\ \hline \mathbf{\dfrac{1}{a_{2012}}+\dfrac{1}{b_{2012}}} &=& \dfrac{1}{a_{1}} + \dfrac{1}{b_{1}}\mathbf{ =\dfrac{1}{2}} \\ \hline \end{array}\)
