數(shù)列的前n項(xiàng)和練習(xí)題

1、數(shù)列的求和訓(xùn)練1.錯(cuò)位相減法求和：如：an等差，bn等比，求aibia2b2anbn的和.1 .求和Sn12x3x2Lnxn11232 .求和：Sn1；二aaa2.裂項(xiàng)相消法求和：把數(shù)列的通項(xiàng)拆成兩項(xiàng)之差、正負(fù)相消剩下首尾若干項(xiàng)。1 -1 .數(shù)列an的前n項(xiàng)和為Sn,若an，則S5等于()n(n1)A.1B.5C.1D.663012.已知數(shù)列an的通項(xiàng)公式為an,求刖n項(xiàng)的和；n(n1)3.已知數(shù)列an的通項(xiàng)公式為，設(shè)Tnaa3,La2a4一1,求Tn.anan21111*4.求111-1,(nN)。121231234123n5.已知等差數(shù)列an滿足a20,a6a810.(1)求數(shù)列an的通項(xiàng)

2、公式及Sna(2)求數(shù)列n1的刖n項(xiàng)和6.已知等差數(shù)列an滿足：a372a726,an的前n項(xiàng)和Sn(1)求an及Sn一一1令bn(nN)，求數(shù)列bn前n項(xiàng)和Tnan1一一17.已知數(shù)列an中，a13,前n和Sn-(n1)(an1)1求證：數(shù)列an是等差數(shù)列求數(shù)列an的通項(xiàng)公式設(shè)數(shù)列1的刖n項(xiàng)和為Tn,是否存在實(shí)數(shù)M,使得TnM對(duì)一切正整數(shù)anan1n都成立？若存在，求M的最小值，若不存在，試說(shuō)明理由_1解：:Sn(n1)(an1)12-1Sm2(n2)(an11)1an1Sn1Sn12(n2)(an11)(n1)(an1)整理得，nan1(n1)an1(n1)an2(n2)an11(n1)a

3、n2nan1(n2)an1(n1)an2(n1)an1(n1)02an)2an1an2an數(shù)列an為等差數(shù)列 a13,nan1(n1電1a22a115a2a12即等差數(shù)列an的公差為2ana1(n1)d3(n1)22n11 1anan1(2n1)(2n3)22n12n311112(3551 11、(-)2 32n3又當(dāng)nN時(shí)，Tn要使得TnM對(duì)一切正整數(shù)n恒成立，只要1 -,所以存在實(shí)數(shù) M使得Tn M對(duì)一切6正整數(shù)n都成立，M的最小值為8.在數(shù)列an中,a11,an(1-)annn12nbn(I)設(shè)an求數(shù)列bn的通項(xiàng)公式(II)求數(shù)列an的前n項(xiàng)和Snan1分析：(I)由已知有n1an1.72nbnbn12n利用累差迭加即可求出數(shù)列b21rinkJn乙n*bn)的通項(xiàng)公式：2(nN)(II)由(I)an知2nn(2kSn_k1'(2k)k1n(2k