(教案)递推数列求通项公式

(教案)递推数列求通项公式 数列复习资料专题一由数列的前几项写通项公式191733?3356399325374 （2）?751381911 （3）1，0，-1，0，1，0，-1，0? （4）1，3，7，15，31?例1 （1）1专题二由S n求a n例2已知数列1，2，4，?的前n项和S n?an3?bn2?.求a n及a，b，c15n2n?1a?,b?0,c?,a n?6622例3设数列?a n?的前n项和为S n，已知a1?a，a n?1?S n?3n设b n?S n?3n，求数列?b n?的通项公式？答依题意，S n?1?S n?a n?1?S n?3n即S n?1?2S n?3n，由此得S n?1?3n?1?2(S n?3n)因此可求通项公式为b n?S n?3n?(a?3)2n?1练习a1?1n(n?1)a n数列?a n?的通项公式？且S n?6210n)（n?N*）试求该数列?a n?有11专题三求数列的最大小项。 例3已知数列?a n?的通项公式a n?(n?1)(没有最大项？若有，求最大项和最大项的项数，若没有，说明理由。 练习已知a n?第几项？专题四递推数列通项公式的求法请牢记以下各种类型的递推数列及a n的求法，考试一般就如下类型。 1a n?1?a n?f(n)（f(n)能够求和）方法累加法法a n?(a n?a n?1)?(a n?1?a n?2)?(a2?a1)?a1n?98*（n?N），则该数列在前30项中，最大项与最小项是n?99例1在数列?a n?中，a1?答案a n?11，a n?1?a n?求数列?a n?的通项公式？24n2?14n?34n?22.a n?1?f(n)a n方法累乘法法a n?a n a n?1a?2?a1?f(n?1)f(n?2)?f (1)a1a n?1a n?2a1例2:在数列?a n?中，a1?1na n?1?2S n,求数列?a n?的通项公式？(提示(n?1)a n?2S n?1)答案a n?n3.a n?1?pa n?q(p,q为常数):方法参数法方法方程组法例3:在数列?a n?中,a1?1,a n?1?2a n?1,求数列?a n?的通项公式？法(参数法)设a n?1?2(a n?)?a n?1?2a n?.对比已知?1?a n?1?1?2(a n?)令b n?a n?1则数列?b n?是以b1?a1?1?2为首项,公比为2的等比数列.?b n?a n?1?b1q n?1?2?2n?1?a n法(方程组法)由a n?1?2a n?1?a n?2a n?1?1?,故?得:a n?1?a n?2(a n?a n?1),这是数列?a n?1?a n?以a2?a1为首项,2为公比的等比数列.4.a n?1?ca n?dnp例4.在数列?a n?中,a1?1,a n?a n?1?3n,求数列?a n?的通项公式？解:由已知a n1a n?1a n1?1b?b?b n?1?1?转化为3类型.,令n n3n33n?13n3?练习:在数列?a n?中,a1?a,且a n?1?2S n?2n?n2,n?N,求数列?a n?的通项公式？n2?a n?1?2S n?2?n解:由?n?12?a n?2S n?1?2?(n?1),(n?2)有a n?1?3a n?2n?1?2n?1n?1法一:(待定系数法)设a n?1?p2?q(n?1)?b?3(a n?p2n?qn?b)则有p?1,q?1,b?02n所以a n?2?n以3为公比,则a n?a,(n?1)?(2a?7)3n?2?2n?1?n,(n?2)a n?1a n?1a n2n?12n?12n?12n?1法二:有n?1?n?n?1?n?1,设b n?1?n?1,由b n?1?b n?n?1?n?13333333用迭带法解之,(注右边当作两数列,等比,与等比差数列,故能求和)5.分式递推数列,一般取”倒”的方法:形式a n?1?ca nba n?d例5.在数列?a n?中,a1?13a n,a n?1?,求数列?a n?的通项公式？22a n?112?b?转化为3类型.n?1n33解:2a?121111?n?,令b n?则有ba n?13a n33a na n6.(第5类型变形)a n?1a n?pa n?1?qa n类型,一般处理为:若p?q,则转化为1111p11?从而为等差数列.若p?q,则可化为?,即转化为类型3.a n?1a np a n?1q a n q例6.已知数列?a n?满足a1?解:由题薏知:1,a na n?1?2a n?a n?1,求数列?a n?的通项公式？2111?(?1),a n?12a n?a111?1?(?1),2a nn?1?111?1是首项为,公比为的等比数列.?1?1?a2a?n?1?2n?111?1?n?1即a n?n?12?1a n22练习:已知数列?a n?满足a1?1,当n?2时,其前n项和S n满足S n?a n(S n?),求数列12?a n?的通项公式？解:当n?2时,S n?(S n?S n?1)(S n?)即2S n S n?1?S n?1?S n2?S?a?1?0,?S1112n?0?S1?S1nn?2,n?11?1?是以2为公差,?1为首项的等差数列.S S?n?1?S1?2n?1,?S n?12n?1当n?2时,a n?S n?S n?1?112?2n?12n?3(2n?1)(2n?3)?1(n?1)?故a n?2?(n?2)?(2n?1)(2n?3)?7（了解）.a n?1?形式.例7.已知数列?a n?满足a1?f(n)a n类型,一般为等式两边取倒数后转化为a n?1?pa n?q的g(n)a n?d(n)33(n?1)a n,a n?1?,求数列?a n?的通项公式？22a n?n设知解:由题1n2?a n?13(n?1)a n3(n?1).即n?11n2n?11n?.设?(?)可求得?1,a n?13a n3a n?13a n?n?1?a?n?是1n1111?1?为首项,为公比的等比数列.则?1?()n?1,得3a n33a13n?3na n?n3?18(了解).对于a n?1?aa n?b类型,一般采用待定系数法,转化为等式两边取倒数,变为ca n?dax?b的根求系数.cx?db n?1?pb n?q的形式.也可用特征方程x?例8:数列?a n?满足a1?1,且8a n?1a n?16a n?1?2a n?5?0记b n?11a n?2,求数列?b n?的通项公式？及数列?a nb n?的前n项和S n?解:法一:由题设知a n?11?,代入递推关系8a n?1a n?16a n?1?2a n?5?0,整b n2理得:4463?0即b n?1?2b n?,3b n?1b nb n?1b n?bnn?1?44?2(b n?).33n?1424?b?是首项为,公比为2的等比数列.故b?2333?n?,即b n?1n(2?4).3a n?11?,b n2n1?a b?1b?1?223n n nn?15?.3?S b?a?na n1b1?a2b2?1(1?2n)51?3?n?(2n?5n?1)1?233点评:若试图从8a n?1a n?16a n?1?2a n?5?0中求a n,进而求b n,将会走进死胡同.将条件b n?11a n?2代入上式,转化为类型3,从而解决.法二:(特征方程)有已知:a n?1?2x?5152a n?5,故x?解有x1?,x2?16?8x2416?8a n156(a n?)12(a n?)142，a?5?即a n?1?n?1416?8a n216?8a n1?111?a n?a?a?1?n2?1222?2为首项,以则相除,有,故数列?是以?5255?a n?5?a n?1?a n?a1?444?4?a n?1?12n?1?5为公比的数列,则a n?n.22?4故a n?13?n,则b n?22?41?(2n?4)13a n?2a n?1?2a n?516?8a n,设1法三.有已知:a n?1?2a n?5?16?8a n,得(2?8?)a n?5?16?(2?8?)(a n?)?5?14?8?2,a n?1?a n?1?16?8a n16?8a n2令5?14?8?0,即?15,?24156(a n?)12(a n?)1142,当?5时,有a?5?当?时,有a n?1?n?124416?8a n216?8a n1?111?a n?a?a?1n2?12,故数列?2?2为首项,以12?是以则相除,有?52552?a n?5?a n?1?a n?a1?444?4?a n?1?2n?1?5为公比的数列,则a n?n2?4例9.在数列?a n?中,a1?2,a n?1?2,求数列?a n?的通项公式？a n?1?(a n?)?2?22解,(带待定系数法)a n?1?a n?1?a n?1a n?12令?2?0得?2或?1.(任选一个算)当?2时,a n?1?2?2(a n?2)1111,化简向类型5转化.?a n?1a n?1?22(a n?2)2111(?2)n?1?2令b n?向类型3转化:b n?1?b n?.再求解.a n?22a n?2(?2)n?1?19a n?1?pa nq(p,a n?0)类型,常用对数转化.?lga n?1?lg p?qlga n令b n?lga n,得b n?1?qb n?lg p转化为3型.例10.在数列?a n?中,a1?1,a2?3,a n?2a n?1a n?2?a n?1a n?1,求数列?a n?的通项公式？a n解:比数列.lg1a n?1?a n?1?lg?lg?是以q?1首项为lga22a n2a1?a n?13()n?121()n?132?lg3的等?a n?1lg?lga n?lg a n?121?na2a3a n202?12?221?n?3?a n?a1?1?3?3?3?3a na1a2a n?1?1?1?2?11?2n?1?320?2?1?2?2?21?n?3.10.(二次一阶递推数列)一般分解因式降次,为能否化生为熟.22例11.在数列?a n?中,a1?1,a n?0,a n?1?a na n?1?2a n.求数列?a n?的通项公式？22在数列?a n?中,a1?1,a n?0,(n?1)a n?1?a na n?1?na n.求数列?a n?的通项公式？二.二阶递推数列.?a1?1,a2?6(p,q为常数).常向等比数列转化,用带待定系数.?a n?2?pa n?1?qa n?0例12.在数列?a n?中,a1?1,a2?5.a n?1?5a n?6a n?1?0,求数列?a n?的通项公式？解:令a n?1?a n?k(a n?a n?1)?a n?1?(k?)a n?k?a n?1或,比较?k?5?2?k?6?k?3?3?a n?1?2a n?3(a n?2a n?1)?k?2或a n?1?3a n?2(a n?3a n?1)均为等比数列.法2.只要一组?a n?1?2a n?3?3n?1?3n,?a n?1?3a n?2?2n?1?2n,两式相减:a n?3n?2n.?2得a n?1?2a n?3(a n?2a n?1),这是以a2?a1为首项3,q?3的等k?3?n?1比数列,?a?2a n?3?3n?1?3n?a n?1?2a n?3n,(用迭代法)作业:在数列?a n?中,a1?1,a2?7,a n?1?4a n?4a n?1?0,求a n(a n?2n?2(3n?1)11(了解周期数列).a n?T?a n,(T?0).对任意的正整数都成立,可利用周期性解决.例13.在数列?a n?中,a1?13,a2?56,对所以的正整数n有:a n?1?a n?a n?2求axx?解,由a n?1?a n?a n?2故a n?2?a n?1?a n?3,两式相加.a n?3?a n,?a n?6?a n?3?a n?a?是以6为周期.?anxx?a334?6?5?a5,经计算axx?56.12归纳猜想型2例14:已知数列?a n?满足a n?1?a n?na n?1,a1?2,求a2,a3,a4猜想a n并证明你的结论.解:由a2?22?1?4?1?52?1?3,a3?32?2?3?1?4,a4?42?3?猜想a n?n?1证明:当n?1时,a1?2?1?1成立假设n?k(k?1,k?N?)时成立,即a k?k?12则n?k?1时,a k?1?a k?ka k?1?(k?1)2?k(k?1)?1?(k?1)?1即n?k?1时也成立.13运用1.已知定义在(0,1)上的函数f(x),对任意的m,n?(1,?)且m?n时,都有11m?n1f()?f()?f().记a n?f (2)n?N?,则数列m n1?mn n?5n?5中,a1?a2?a8?C Af()B f()C f()D f()分析:a n?f(?a n?121314151(n?2)?(n?3)11)?f()?f()?f(),再由迭加法.2n2?5n?51?(n?2)(n?3)n?2n?33x,(x?R),正项等比数列?a n?满足a50?1,则2.已知函数f(x)?x3?1f(lna1)?f(lna2)?f(lna99)?C A99B101C分析:99101D22因为f(?x)?f,?且xf(lna99)?f(ln1)?f(?lna1)即f(lna99)?f(lna1)?1.a1xx5a n?1?2?3数列?a n?满足a n?(n?2,n?N),数列的前xx项的和为403,则?a ia i?1a n?1?5i?1的值是A A10B9C8D7分析(考查数列的周期性与转化思想)设a1?a,a2?由a n?xx5a?2.求得a3?a所以数列周期为2a?55a n?1?2?a n?1a n?5(a n?1?a n)?2a n?1?5?5?(a i?a i?1)?2?xx?5?2?a i?2?xx?10i?1i?1xxxx?a ai?1i i?14.已知两个等差数列?a n?，?b n?的前n项和S n，T n，且41S n3n?2a，则7?T n2n?1b51937S n2n?3a，则9?T n3n?1b9505已知两个等差数列?a n?，?b n?的前n项和S n，T n，且注意3,4的解法是有区别的6已知数列?a n?的前n项和S n满足，3S n?4a n?2n?4,n?N?， （1）证明当n?2时，a n?4a n?1?2 （2）求数列的?a n?的通项公式。 （3）设c n?2n?1a n,T n为数列?c n?的前n项和，证明T n?8a n?1解 （1）由3Sn?4a n?2n?4，得当n?2时，3S n?1?4a n?1?2(n?1)?4,。 则有3(S n?S n?1)?4a n?4a n?1?2，故a n?4a n?1?2 （2）当n?2时，a n?4a n?1?2?a n?23?4(a n?1?)所以数列?a n?2是以a1?2为首项，4为公比的等比数列。 33?又3S1?4a1?2?a1?2?a1?232324?。 334n?14n4n?2?a n?,n?N?则a n?4?333a n4n?24n?212 （3）c n?n?1?n?1?n?1a n?14?2444当n?1时,T1?c1?a113?a238当当n?2时，T n?c1?c2?c3?c n?a1n?1111?2(3?4?n?1)a24444111?3n?11n?1444?2?1341?42n?122n?1?即证:83?4n?18?11?7.已知数列?a n?的前n项和S n,m?(S n,n?1),n?(,n),且a n?1?m?n,a1?1.求n2a n的通项公式.S n n?1S n?1?n?S n?1?S n?n?nn2n2解:S S1?n?1?n?nn?1n2a n?1?S S SSS2S1S2S1?(?)?(3?2)?(n?n?1)212132n n?1所以:1111n?1?2?n?1?2?1?()?2222?S n n?1n?1?n?2?nn22n?2当n?2时,a n?2?n?1,又当n?1时,a1?12n?2所以a n?2?n?1228.已知数列?a n?的前n项和S n(n?N?),a1?,且当n?2时.S nS n?1?3S n?2?03则a n?1? (1)求a2,a3的值 (2)若b n?1,求数列?b n?的通项公式S n?1?Sn?1?1?n1n?T?的前项和,证明T n?n nS?1272?n? (3)设数列?法一(猜想归纳法)248,易知a2?,a3?32110526143062,S4?,S5? (2)由 (1)知:S1?,S2?,S3?37153163 (1)由a1?2n?1?2猜想:S n?n?12?1证明:当n?1时,显然成立.2k?1?2假设n?k时也成立,即S k?k?1,当n?k?1时,由S k?1S k?3S k?1?2?02?122(k?1)?1?2知S k?1?即证.?(k?1)?13?S k2?1则S n?1?11,故b?2n?1?1nn?12?1S n?1 (3)令c n?S n?1?12?11?S n?12n?2?122n?2?1n?1121n则T n?c1?c2?c3?c n?即证22111117118111?又c n?222n?2?12142n?2?12142n?2272n所以c n?所以T n?c1?c2?c3?c n?n1?112131?()?()?()n?27?2222?1?1n?1?()?n12?2?n1?=?即证.127271?2法二 (1)略 (2)当n?2时.S nS n?1?3S n?2?0即S n?23?S n?1则S n?1?S?12?1?n?13?S n?13?S n?1有3?S n?1121?1?而b n?S n?1S n?1?1