新高考数学二轮复习 数列重难点提升专题04 构造法求数列通项的八种技巧(一)（原卷版）

专题04构造法求数列通项的八种技巧(一)【必备知识点】◆构造一:待定系数之SKIPIF1<0型构造等比数列求关于SKIPIF1<0(其中SKIPIF1<0均为常数,SKIPIF1<0)类型的通项公式时,先把原递推公式转化为SKIPIF1<0,再利用待定系数法求出SKIPIF1<0的值,再用换元法转化为等比数列求解.其实对于这类式子,我们只需要记住在等式两侧加上一个常数SKIPIF1<0,构造成等比数列.常数SKIPIF1<0的值并不需要背诵,我们可以通过待定系数法推导出来.【经典例题1】已知SKIPIF1<0满足SKIPIF1<0,SKIPIF1<0求数列SKIPIF1<0的通项公式.【经典例题2】已知数列SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,求数列SKIPIF1<0的通项公式.【经典例题3】已知数列SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,求数列SKIPIF1<0的通项公式.【练习1】数列SKIPIF1<0中,SKIPIF1<0,设其前SKIPIF1<0项和为SKIPIF1<0,则SKIPIF1<0A.SKIPIF1<0B.SKIPIF1<0C.15D.27【练习2】已知数列SKIPIF1<0的前SKIPIF1<0项和为SKIPIF1<0,若SKIPIF1<0,则SKIPIF1<0A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【练习3】在数列SKIPIF1<0中,SKIPIF1<0,则SKIPIF1<0_______.【练习4】已知数列SKIPIF1<0满足SKIPIF1<0,则数列SKIPIF1<0的通项公式SKIPIF1<0=______.【练习5】已知数列SKIPIF1<0的首项SKIPIF1<0,且SKIPIF1<0,则数列SKIPIF1<0的前10项的和为______.【练习6】已知数列SKIPIF1<0中,SKIPIF1<0,则SKIPIF1<0_______.◆构造二:待定系数之SKIPIF1<0型构造等比数列求关于SKIPIF1<0类型的通项公式时,与上面讲述的构造一的方法很相似,只不过等式中多了一项SKIPIF1<0,在构造时我们也保持跟题干一样的结构,加一项SKIPIF1<0再构造等比数列就可以,即令SKIPIF1<0,然后与已知递推式各项的系数对应相等,解SKIPIF1<0,从而得到SKIPIF1<0是公比为SKIPIF1<0的等比数列.【经典例题1】设数列SKIPIF1<0满足SKIPIF1<0,SKIPIF1<0,求数列SKIPIF1<0的通项公式.【经典例题2】已知:SKIPIF1<0,SKIPIF1<0时,SKIPIF1<0,求SKIPIF1<0的通项公式.【练习1】已知数列SKIPIF1<0是首项为SKIPIF1<0.(1)求SKIPIF1<0通项公式;(2)求数列SKIPIF1<0的前SKIPIF1<0项和SKIPIF1<0.【练习2】已知数列SKIPIF1<0和SKIPIF1<0的前SKIPIF1<0项和SKIPIF1<0,对于任意的SKIPIF1<0是二次方程SKIPIF1<0SKIPIF1<0的两根.(1)求SKIPIF1<0和SKIPIF1<0通项公式;(2)SKIPIF1<0的前SKIPIF1<0项和SKIPIF1<0.【练习3】设数列SKIPIF1<0是首项为SKIPIF1<0,满足SKIPIF1<0.问是否存在SKIPIF1<0,使得数列SKIPIF1<0成等比数列?若存在,求出SKIPIF1<0的值,若不存在,说明理由;◆构造三:待定系数之SKIPIF1<0型构造数列求关于SKIPIF1<0(其中SKIPIF1<0均为常数,SKIPIF1<0)类型的通项公式时,共有3种方法.方法一:先用待定系数法把原递推公式转化为SKIPIF1<0,根据对应项系数相等求出SKIPIF1<0的值,再利用换元法转化为等比数列求解.方法二:先在递推公式两边同除以SKIPIF1<0,得SKIPIF1<0,引入辅助数列SKIPIF1<0(其中SKIPIF1<0SKIPIF1<0),得SKIPIF1<0,再利用待定系数法解决；方法二:也可以在原递推公式两边同除以SKIPIF1<0,得SKIPIF1<0,引入辅助数列SKIPIF1<0(其中SKIPIF1<0),得SKIPIF1<0SKIPIF1<0,再利用叠加法(逐差相加法)求解.【经典例题1】已知数列SKIPIF1<0中SKIPIF1<0,求SKIPIF1<0的通项公式.【经典例题2】已知数列SKIPIF1<0满足SKIPIF1<0,求数列SKIPIF1<0的通项公式.【练习1】已知数列SKIPIF1<0满足SKIPIF1<0.设SKIPIF1<0,若对于SKIPIF1<0,都有SKIPIF1<0恒成立,则SKIPIF1<0的最大值为()A.3B.4C.7D.9【练习2】已知数列SKIPIF1<0满足SKIPIF1<0.(1)判断数列SKIPIF1<0是否为等差数列,并说明理由;(2)记SKIPIF1<0为数列SKIPIF1<0的前SKIPIF1<0项和,求SKIPIF1<0.【过关检测】一、单选题1．已知SKIPIF1<0为数列SKIPIF1<0的前n项和，若SKIPIF1<0，则SKIPIF1<0的通项公式为（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．已知数列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，则数列SKIPIF1<0的通项公式为（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．已知数列SKIPIF1<0满足SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0的值为（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．设数列SKIPIF1<0的前n项和为SKIPIF1<0，若SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．在数列SKIPIF1<0中，SKIPIF1<0，且SKIPIF1<0，则SKIPIF1<0的通项为（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<06．数列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．数列SKIPIF1<0满足SKIPIF1<0，且SKIPIF1<0，若SKIPIF1<0，则SKIPIF1<0的最小值为（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<08．已知数列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），则数列SKIPIF1<0通项公式SKIPIF1<0为（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．数列SKIPIF1<0满足SKIPIF1<0且SKIPIF1<0，则此数列第5项是（

）A．15 B．255 C．16 D．6310．在数列SKIPIF1<0中，已知SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<011．在数列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，则n的最小值是（

）A．8 B．9 C．10 D．1112．设数列{an}中，a1＝2，an＋1＝2an＋3，则通项an可能是（）A．5－3n B．3·2n－1－1C．5－3n2 D．5·2n－1－313．在数列SKIPIF1<0中，若SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<014．已知在数列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<015．数列SKIPIF1<0满足SKIPIF1<0，若SKIPIF1<0，则SKIPIF1<0的取值范围为（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、填空题16．设数列SKIPIF1<0满足SKIPIF1<0，且SKIPIF1<0，则数列SKIPIF1<0的通项公式为SKIPIF1<0___________.17．已知数列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0通项SKIPIF1<0______；18．数列{an}满足a1＝1，an＋1＝2an＋1.(n∈N*).数列{an}的通项公式为______．19．数列SKIPIF1<0满足SKIPIF1<0，且SKIPIF1<0，则SKIPIF1<0_________.20．已知数列SKIPIF1<0满足SKIPIF1<0，且SKIPIF1<0前8项和为761，则SKIPIF1<0______．三、解答题21．已知数列SKIPIF1<0满足SKIPIF1<0.(1)证明SKIPIF1<0为等比数列，并求SKIPIF1<0的通项公式；(2)记数列SKIPIF1<0的前SKIPIF1<0项和为SKIPIF1<0，证明SKIPIF1<0