觀察下列各式:a+b=1,a2+b2=3,a3+b3=4,a4+b4=7,a5+b5=11,…,則a10+b1...

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問題詳情:

觀察下列各式:a+b=1,a2+b2=3,a3+b3=4,a4+b4=7,a5+b5=11,…,則a10+b1...

觀察下列各式:ab=1,a2+b2=3,a3+b3=4,a4+b4=7,a5+b5=11,…,則a10+b10=(  )

A.121                              B.123

C.231                                       D.211

【回答】

B.法一:令ananbn,則a1=1,a2=3,a3=4,a4=7,…,得an+2=anan+1,從而a6=18,a7=29,a8=47,a9=76,a10=123.

法二:由ab=1,a2+b2=3,得ab=-1,代入後三個等式中符合,則a10+b10=(a5+b5)2-2a5b5=123.

知識點:推理與*

題型:選擇題

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