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{an=an−1an−2−bn−1bn−2bn=bn−1an−2+an−1bn−2 \large \begin{cases} {a_n=a_{n-1}a_{n-2}-b_{n-1}b_{n-2} } \\ { b_n=b_{n-1}a_{n-2}+a_{n-1}b_{n-2} } \end{cases} ⎩⎨⎧an=an−1an−2−bn−1bn−2bn=bn−1an−2+an−1bn−2
Two sequences ana_nan and bnb_nbn begin with values a1=b1=a2=b2=1a_1=b_1=a_2=b_2=1a1=b1=a2=b2=1 and satisfy the recursive relations above for n>2n>2n>2.
Find b2015a2015\large \frac{b_{2015}}{a_{2015}}a2015b2015.
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