已知數(shù)列遞推公式求通項(xiàng)公式的幾種方法解析

求數(shù)列通項(xiàng)公式的方法一、公式法aaa32n，a=2，求數(shù)列{a}的通項(xiàng)公式。nn+1n1naa3aa3a解：a=2a+32n兩邊除以2n+1，得n+1=n+，則n+1一n=，故數(shù)列{aa3aa3an+1n2n+12n22n+12n22n21222n2nn22aa3評注：本題解題的關(guān)鍵是把遞推關(guān)系式a=2a+32n轉(zhuǎn)化為n+1一n=，說明數(shù)列n+1n2n+12n22n2n2{a}的通項(xiàng)公式。n二、累加法aaa+2n+1，a=1，求數(shù)列{a}的通項(xiàng)公式。nn+1n1nn+1nn+1nnnnnn322112=n2所以數(shù)列{a}的通項(xiàng)公式為a=n2。nnaann+1nn+1nnnnn11n3n22 n22 例3已知數(shù)列{a}滿足a=a+23n+1，a=3，求數(shù)列{a}的通項(xiàng)公式。nn+1n1nn+1nn+1nnnnnn32211 nn+1nn+1nnnnnn211naaa23n+1，a=3，求數(shù)列{a}的通項(xiàng)公式。nn+1n1naa21解：a=3a+23n+1兩邊除以3n+1，得n+1=n++，n+1n3n+13n33n+1aa21則n+1一n=+，故3n+13n33n+11=(+)+(+)+=(+)+(+)+(+)++(+)+n3n一133n一23323211n+1naa21aa21nn3n+13n33n+13n3n一13n一13n一23n一23n一332313l3nJ的通項(xiàng)公式，最后再求數(shù)列{a}的通項(xiàng)公式。n三、累乘法例5已知數(shù)列{a}滿足a=2(n+1)5n〉a，a=3，求數(shù)列{a}的通項(xiàng)公式。nn+1n1nn+1n1nanaaaaa=n.n一1..3.2.anaaaa1nn評注：本題解題的關(guān)鍵是把遞推關(guān)系a=2(n+1)5n〉a轉(zhuǎn)化為n+1=2(n+1)5n，進(jìn)而求n+1nanaaaa出n.n一1..3.2.a，即得數(shù)列{aaaaaaaaa1nnn1n123n一1nn123n3n一1n進(jìn)而求出n.進(jìn)而求出n.n1.ananaa2222222③由a=a+2a+3a++(n1)a(n2)，取n=2得a=a+2a，則a=a，又知n123n121221a=1，則a=1，代入③得a=1.3.4.5..n=。2n2nn2.nn2.nn+1nanaa2 四、待定系數(shù)法nn+1n1n解：設(shè)a+x5n+1=2(a+x5n)④n+1n將a=2a+35n代入④式，得2a+35n+x5n+1=2a+2x5n，等式兩邊消去n+1nnnnn+1naa5n0，則=2，則數(shù)列{a5n}是以na5nnn1nnaana5n+1=2(a5n)，n+1nn+1n從而可知數(shù)列{a5n}是等比數(shù)列，進(jìn)而求出數(shù)列{a5n}的通項(xiàng)公式，最后再求出數(shù)列nn{a}的通項(xiàng)公式。n例8已知數(shù)列{a}滿足a=3a+52n+4，a=1，求數(shù)列{a}的通項(xiàng)公式。nn+1n1n解：設(shè)a+x2n+1+y=3(a+x2n+y)⑥n+1n將a=3a+52n+4代入⑥式，得n+1nnn(5+2x=3x(x=5yyyy=2a+52n+1+2=3(a+52n+2)n+1n⑦1nn1nn評注：本題解題的關(guān)鍵是把遞推關(guān)系式a=3a+52n+4轉(zhuǎn)化為n+1na+52n+1+2=3(a+52n+2)，從而可知數(shù)列{a+52n+2}是等比數(shù)列，進(jìn)而求n+1nn出數(shù)列{a+52n+2}的通項(xiàng)公式，最后再求數(shù)列{a}的通項(xiàng)公式。nn例9已知數(shù)列{a}滿足a=2a+3n2+4n+5，a=1，求數(shù)列{a}的通項(xiàng)公式。nn+1n1n解：設(shè)a+x(n+1)2+y(n+1)+z=2(a+xn2+yn+z)⑧n+1n將a=2a+3n2+4n+5代入⑧式，得n+1nnnnnnn+1nnn1nn評注：本題解題的關(guān)鍵是把遞推關(guān)系式a=2a+3n2+4n+5轉(zhuǎn)化為nnn+1nnn求出數(shù)列{a}的通項(xiàng)公式。n五、對數(shù)變換法例10已知數(shù)列{a}滿足a=2人3n人a5，a=7，求數(shù)列{a}的通項(xiàng)公式。nn+1n1nn+1n1nn+1n+1n常用對數(shù)得lga=5lga+nlg3+lg2⑩n+1nn+1nnnn((lg3+x=5x|x=lg34n+14164n4164141644164n4164n4164n41644164n41644164n4164464111n11n評注：本題解題的關(guān)鍵是通過對數(shù)變換把遞推關(guān)系式a=2〉3n〉a5轉(zhuǎn)化為n+1n.lga3.lga2.lga3.lga2.lga=lg53n1.n!.2n(1)，從而a=53n1.n!.2lgalga1nlgnlglglgalgnlglgn+14164n4164{lga+lg3n+lg3+lg2}是等比數(shù)列，進(jìn)而求出數(shù)列{lga+lg3n+lg3+lg2}的通項(xiàng)n4164n4164公式，最后再求出數(shù)列{a}的通項(xiàng)公式。n六、迭代法nn+1n1nn+1nnn1n2an2).2n3]32(n1).n.2(n2)+(n1)=nnn 1 =a3n1.n!.221nn又a1=5，所以數(shù)列{a}的通項(xiàng)公式為a=53n1.n!.2n(1)。nnn+1n兩邊取常用對數(shù)得lga=3(n+1)2nlga，即=3(n+1)2n，再由累乘法可推知n+1nlganlga=lgan.lgan1.nlgalgan1n2七、數(shù)學(xué)歸納法21例12已知數(shù)列{a}滿足a=a+8(n+1)，a=8，求數(shù)列{a}的通項(xiàng)公式。8(n+1)8解：由a=a+及a=，得n+1n(2n+1)2(2n+3)219a=5a=a+=+=a=a+=+=由此可猜測a=(2n+1)21，往下用數(shù)學(xué)歸納法證明這個(gè)結(jié)論。(21+1)218 (1)當(dāng)n=1時(shí)，a==，所以等式成立。(21+1)29(2)假設(shè)當(dāng)n=k時(shí)等式成立，即a=(2k+1)21，則當(dāng)n=k+1時(shí)，k=+==當(dāng)n=k+1時(shí)等式也成立。根據(jù)(1)，(2)可知，等式對任何n