常見遞推數(shù)列通項(xiàng)的九種求解方法

高考中的遞推數(shù)列求通項(xiàng)問題，情境新穎別致，有廣度，創(chuàng)新度和深度，是高考的熱點(diǎn)之一。是一類考查思求n+1nn1nn-1nn-1：〈|a4-a3=5上述n-1個等式相加可得：nn-1nn-1n1n【類型一專項(xiàng)練習(xí)題】1nn-1nn1n+1nnnn+1n1nnn+1n1nn1n+1nnn1nn-1n7、若數(shù)列的遞推公式為a=3,a=a-2.3n+1(n=N*)，則求這個數(shù)列的通項(xiàng)公式n+1nnnn1nnnn1nn12n+1nn2+nnn1n+1n123(I)求c的值；(II)求{a}的通項(xiàng)公式．nn2nn9.a=_n2n9.a=_n2nn2a=f(n).an+1n (f(n)可以求積)n1n_1nnaaaaa解析：a=n.n_1.n_23.2.anaaaaa1n_1n_2n_321評注：一般情況下，累積法里的第一步都是一樣的?！绢愋投ｍ?xiàng)練習(xí)題】nn1nn+1nnn1n+1nnnn+1n1nnn+1n1nn1n123n_1n9、設(shè){a}是首項(xiàng)為1的正項(xiàng)數(shù)列,且(n+1)a2-na2+a·a=0(n=1,2,3,…)，求它的通項(xiàng)公式.nn+1nn+1nnn1nnnn41nnnnn1nn一1nn【類型三專項(xiàng)練習(xí)題】n1n+1nn2、若數(shù)列的遞推公式為a=1,a=2a一2(neN*)，則求這個數(shù)列的通項(xiàng)公式n+1nn1n2n一1nnn+12n13nn1n+1nnan1n+1nn7、設(shè)二次方程ax2-ax+1=0(n∈N)有兩根α和β，且滿足6α-2αβ+6β=3．nn＋1.nn+1(2)(2)(3)當(dāng)a=7時(shí)，求數(shù)列{a}的通項(xiàng)公式nnn122n+1nn一1nn答案：1.a=3n一22.a=2一2n一13.a=2一21一n4.a=4一.21一n5.a=nnnn3n2nnn一1還原到(*)式，則數(shù)列{a+aa}是以a+aa為首項(xiàng)，b為公比的等比數(shù)列，然后再結(jié)合其它方法，就可n+1n21Ann12n+1nn一1n解析：令a+aa=b(a+aa),(n之2)n+1nnn一1n+1nnn一1n+1n21n+1nnnn一1A+B+C=0,則一定可以構(gòu)造{a一a}為等比數(shù)列。n+1n12n+1n一1nnannnn一1n+1nn一1n+1n212n+122n4nn2nnn+12n4n+12nn+12n2【類型四專項(xiàng)練習(xí)題】n12n+23n+13nn552123n+23n+13nnnnnn+1n1⑴設(shè)數(shù)列b=a-2a(n=1,2,^^)，求證：數(shù)列是等比數(shù)列；nn+1nnn2nnnnnnn+2n+1n12nppn+1n一般需一次或多次待定系數(shù)法，構(gòu)造新的等差數(shù)列或等比數(shù)列。an1n2n-1nn2n一1nn：是以3為首項(xiàng)，以1為公比的等比數(shù)列n2，：nnn一1n2nn2nnn1nn一1nnnnn一1nnnn一1nnnn1nnnn1nn+1nnn+1nn+1nn+1nn+1n2n+122n2a+235令n=b，則b一b=2nnn+12n2n+12nn+12n222n+12nn1222習(xí)題：nn3n33nn12n+1nnn+1nn(2)求數(shù)列{a}的通項(xiàng)；n1n+1nnnnn一1nn1n+1nnn1n+1nnn12nn一1nn6n+13n2n8、已知數(shù)列{a}，a=1,n∈N，a=2a＋3n,求通項(xiàng)公式a．n1+n+1nnnnn1nnnn1n10、若數(shù)列的遞推公式為a=1,a=3a一2.3n+1(n=N)，則求這個數(shù)列的通項(xiàng)公式1n+1n+n1n+1nnnnn1nnnn1nnnn1nnnn1nn1nn一1nnnn+1n1nn+1nnn2nnnnn2nnnn3nnnn 11a2aann+12nn+1nnnnb一22nn1a421評注：去倒數(shù)后，一般需構(gòu)造新的等差(比)數(shù)列。1、若數(shù)列的遞推公式為a=3,=一2(n從￥)，則求這個數(shù)列的通項(xiàng)公式。aan+1nnn一1nn一1nn3、已知數(shù)列{a}滿足：nnn3a+6n5、已知數(shù)列{a}滿足nn3a+6nn7、若數(shù)列{a}中，a=1，a=2ann∈N，求通項(xiàng)a．n1n+1a+2