高中數(shù)學(xué)必修五數(shù)列求和方法總結(jié)附經(jīng)典例題和答案詳解

1、數(shù)列專(zhuān)項(xiàng)之求和數(shù)列專(zhuān)項(xiàng)之求和44（一）等差等比數(shù)列前n項(xiàng)求和1、等差數(shù)列求和公式：dnnnaaanSnn2)1(2)(11?????2、等比數(shù)列求和公式：?????????????)1(11)1()1(111qqqaaqqaqnaSnnn（二）非等差等比數(shù)列前n項(xiàng)求和⑴錯(cuò)位相減法錯(cuò)位相減法②數(shù)列為等差數(shù)列，數(shù)列為等比數(shù)列，則數(shù)列的求和就要采??na??nb??nnab?用此法.②將數(shù)列的每一項(xiàng)分別乘以的公比，然后在錯(cuò)位相減，進(jìn)而可得到數(shù)

2、??nnab???nb列的前項(xiàng)和.??nnab?n此法是在推導(dǎo)等比數(shù)列的前項(xiàng)和公式時(shí)所用的方法.n例23.23.求和：132)12(7531???????????nnxnxxxS)0(?x例24.24.求數(shù)列前n項(xiàng)的和.??????2226242232nn⑵裂項(xiàng)相消法裂項(xiàng)相消法一般地，當(dāng)數(shù)列的通項(xiàng)時(shí)，往往可12()()ncaanbanb???12(abbc為常數(shù)）將變成兩項(xiàng)的差，采用裂項(xiàng)相消法求和.na可用待定系數(shù)法進(jìn)行裂項(xiàng)：設(shè)，通分整

3、理后與原式相比較，根據(jù)對(duì)應(yīng)項(xiàng)系數(shù)相等得12naanbanb??????，從而可得21cbb???12211211=().()()()ccanbanbbbanbanb??????例30.30.求的值?????89sin88sin3sin2sin1sin22222????????⑸記住常見(jiàn)數(shù)列的前記住常見(jiàn)數(shù)列的前項(xiàng)和：項(xiàng)和：n①(1)123...2nnn??????②2135...(21)nn??????③22221123...(1)(21

4、).6nnnn???????④23333)]1(21[321??????nnn?答案詳解答案詳解例23.解：由題可知，的通項(xiàng)是等差數(shù)列2n－1的通項(xiàng)與等比1)12(??nxn數(shù)列的通項(xiàng)之積。1?nx……………………….①132)12(7531???????????nnxnxxxS設(shè)……………………….②（設(shè)制錯(cuò)nnxnxxxxxS)12(7531432??????????位）①－②得（錯(cuò)位相nnnxnxxxxxSx)12(222221)