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本帖最后由 王守恩 于 2020-10-19 12:56 编辑
这样也行!
\(5a_1+8a_2=120\ \ \ (1)\)
\(8a_1+13a_2=?\ \ \ \ (2)\)
\((1)×13/8:8.125a_1+13a_2=195\ \ \ \ (3)\)
\((1)×21/13:8.1a_1+12.9a_2=193.8\ \ (4)\)
\(\displaystyle比较(2),(3),(4):120×\frac{13}{8}>\ \ ?\ >120×\frac{21}{13}\)
\(得:\ \ ?\ =194\)
\(\displaystyle一般地,已知a_{k},求a_{k+1},可以这样:\)
记{F\(_{k}\)}\(_{k=0}^{\infty}=0,1,1,2,3,5,8,13,21,......\)
\(\displaystyle a_{k}*\frac{F_{k}}{F_{k-1}}>a_{k+1}>a_{k}*\frac{F_{k+1}}{F_{k}}\)
\(\displaystyle a_{k+1}的数量=\frac{a_{k}}{F_{k}*F_{k-1}}个(不是唯一解)\)
\(譬如:\displaystyle a_{7}=2358,求a_{8},\ \ a_{8}可以有\frac{2358}{13*8}=22个解\)
\(a_{10}=12358,a_{11}=19991,19992,19993,19994,19995,19996,19997\) |
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