專題05構造法求數列通項的八種技巧(二)【必備知識點】◆構造四:同型構造法所謂同型構造法,就是將找因式中的因子和數列項數相同或者相近的部分通過同除或同乘化歸成結構相同的形式,形成新的數列,如常數列,等差數列或等比數列.下面讓我們來看看有哪些模型結構吧.模型一:SKIPIF1<0,構造SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0為常數數列.模型二:SKIPIF1<0,構造SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0為常數數列.模型三:SKIPIF1<0,構造SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0為常數數列.模型四:SKIPIF1<0,構造SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0為等比數列.模型五:SKIPIF1<0,構造SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0為等比數列.模型六:SKIPIF1<0,構造SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0為等差數列.模型七:SKIPIF1<0,構造SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0為等差數列.模型八:SKIPIF1<0,構造SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0為等差數列.看了這么多模型,是不是覺得很多,很難記住呢,其實向大家展示這么多,只是想向大家展示,當看到這類式子,盡量將SKIPIF1<0和SKIPIF1<0,SKIPIF1<0和SKIPIF1<0等因子和數列項數相同的部分劃歸成結構相同的形式,構造成新數列.【經典例題1】已知數列SKIPIF1<0滿足SKIPIF1<0,求SKIPIF1<0.【經典例題2】已知數列SKIPIF1<0中,SKIPIF1<0且SKIPIF1<0,求數列SKIPIF1<0的通項公式.【經典例題3】已知數列SKIPIF1<0中,SKIPIF1<0且SKIPIF1<0,求數列SKIPIF1<0的通項公式.【經典例題4】已知SKIPIF1<0,且SKIPIF1<0,求數列SKIPIF1<0的通項公式.【練習1】已知數列SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0A.28B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【練習2】已知SKIPIF1<0是首項為1的正項數列,且SKIPIF1<0,則它的通項公式是SKIPIF1<0______________.【練習3】已知數列SKIPIF1<0滿足SKIPIF1<0,求數列SKIPIF1<0的通項公式.【練習4】已知數列SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,求數列SKIPIF1<0的通項公式.【練習4】已知數列SKIPIF1<0前SKIPIF1<0項的和為SKIPIF1<0,且滿足SKIPIF1<0.(1)求SKIPIF1<0的值;(2)求SKIPIF1<0的通項公式.◆構造五:取倒數構造等差類型一:數列SKIPIF1<0滿足:SKIPIF1<0,則有SKIPIF1<0.所以SKIPIF1<0是以SKIPIF1<0為首項,SKIPIF1<0為公差的等差數列,即SKIPIF1<0.(當分母出現加減時,我們很難將它進行化簡運算,所以往往取倒數再運算才能找到突破點).類型二:數列SKIPIF1<0滿足:SKIPIF1<0,則有SKIPIF1<0.所以SKIPIF1<0是等差數列.類型三:若數列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0,且滿足SKIPIF1<0,則有SKIPIF1<0,兩邊同除以SKIPIF1<0得:SKIPIF1<0,故SKIPIF1<0是以SKIPIF1<0為首項,SKIPIF1<0為公差的等差數列,即SKIPIF1<0,再用SKIPIF1<0,求SKIPIF1<0.【經典例題1】在數列SKIPIF1<0中,若SKIPIF1<0,則SKIPIF1<0___________.【經典例題2】已知數列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0,且滿足SKIPIF1<0.(1)求證:SKIPIF1<0是等差數列.(2)求SKIPIF1<0的表達式.【經典例題3】已知數列SKIPIF1<0的首項SKIPIF1<0,證明:數列SKIPIF1<0是等比數列并求SKIPIF1<0的通項公式.【練習1】設SKIPIF1<0是數列SKIPIF1<0的前SKIPIF1<0項和,且SKIPIF1<0,則SKIPIF1<0A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【練習2】已知SKIPIF1<0中,SKIPIF1<0,則SKIPIF1<0_______.【練習3】已知數列SKIPIF1<0,求SKIPIF1<0的通項公式.【過關檢測】一、單選題1．已知數列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則數列SKIPIF1<0的前100項的和是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．數列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則下列結論錯誤的是（

）A．SKIPIF1<0 B．SKIPIF1<0是等比數列C．SKIPIF1<0 D．SKIPIF1<03．若數列SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．已知數列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則滿足SKIPIF1<0的n的最大取值為（

）A．7 B．8 C．9 D．105．已知數列SKIPIF1<0滿足SKIPIF1<0，則滿足SKIPIF1<0的SKIPIF1<0的最大取值為（

）A．6 B．7 C．8 D．96．已知數列SKIPIF1<0滿足：SKIPIF1<0，SKIPIF1<0SKIPIF1<0，則SKIPIF1<0(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．已知數列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<08．已知數列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，則數列SKIPIF1<0的通項公式為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、多選題9．設數列SKIPIF1<0的前n項和為SKIPIF1<0，已知SKIPIF1<0，且SKIPIF1<0SKIPIF1<0，則下列結論正確的是（

）A．SKIPIF1<0是等比數列 B．SKIPIF1<0是等比數列C．SKIPIF1<0 D．SKIPIF1<010．已知數列{SKIPIF1<0}滿足SKIPIF1<0，SKIPIF1<0，則下列結論正確的是（

）A．SKIPIF1<0為等比數列 B．{SKIPIF1<0}的通項公式為SKIPIF1<0C．{SKIPIF1<0}為遞增數列 D．SKIPIF1<0的前n項和SKIPIF1<0三、填空題11．已知數列SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0___________.12．已知數列SKIPIF1<0滿足SKIPIF1<0，則數列SKIPIF1<0的前SKIPIF1<0項和為______．13．已知數列SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，則數列SKIPIF1<0__________14．已知數列SKIPIF1<0的首項SKIPIF1<0，且對任意的SKIPIF1<0，都有SKIPIF1<0，則SKIPIF1<0______．15．已知數列SKIPIF1<0滿足：SKIPIF1<0，SKIPIF1<0（SKIPIF1<0N+），由SKIPIF1<0、SKIPIF1<0、SKIPIF1<0歸納出數列SKIPIF1<0的通項公式是__________.16．數列SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0___________.17．已知數列SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，則數列|SKIPIF1<0的前n項和為______