题目：
观察下列各式：a＋b＝1；a2＋b2＝3；a3＋b3＝4；a4＋b4＝7；a5＋b5＝11；…；则a10＋b10＝________．
答案：
123
(解法1)由a＋b＝1；a2＋b2＝3得ab＝－1代入后三个等式中符合，则a10＋b10＝(a5＋b5)2－2a5b5＝123.
(解法2)令an＝an＋bn，易得an＋2＝an＋an＋1，从而a6＝18，a7＝29，a8＝47，a9＝76，a10＝123.
观察下列各式：a＋b＝1；a2＋b2＝3；a3＋b3＝4；a4＋b4＝7；a5＋
来源：www.tikuol.com
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