高三数学复习第六章数列第四节数列求和理.pptx

,教材研读,考点突破,栏目索引,教材研读,考点突破,栏目索引,教材研读,考点突破,栏目索引,*,*,教材研读,考点突破,栏目索引,*,*,教材研读,考点突破,栏目索引,教材研读,考点突破,栏目索引,*,*,理数,课标版,第四节数列求和,1/23,1.求数列前,n,项和方法,(1)公式法,(i)等差数列前,n,项和公式,S,n,=,=,na,1,+,.,教材研读,(ii)等比数列前,n,项和公式,当,q,=1时,S,n,=,na,1,;,当,q,1时,S,n,=,=,.,2/23,(2)分组转化法,把数列每一项转化成几项之和,使所求和转化为几个等差、等比数列,之和,再求解.,(3)裂项相消法,把数列通项拆成两项之差再求和,正负相消剩下首尾若干项.,(4)倒序相加法,把数列分别正着写和倒着写再相加,倒序相加法是对等差数列求和公式,推导过程推广.,(5)错位相减法,适合用于一个等差数列与一个等比数列对应项相乘所得数列求和,错,位相减法是对等比数列求和公式推导过程推广,.,3/23,(6)并项求和法,求一个数列前,n,项和时,可将数列项合并求解,这种方法称为并项求,和.形如,a,n,=(-1),n,f,(,n,)类型,可采取两项合并求和.比如:,S,n,=100,2,-99,2,+98,2,-,97,2,+,+2,2,-1,2,=(100+99)+(98+97)+,+(2+1)=5 050.,2.常见裂项公式,(1),=,-,;,(2),=,;,(3),=,-,.,4/23,1.数列,a,n,通项公式是,a,n,=,前,n,项和为9,则,n,等于,(),A.9B.99C.10D.100,答案,B,a,n,=,=,-,S,n,=,a,1,+,a,2,+,+,a,n,=(,-,)+,(,-,)+,+(,-,)+(,-,)=,-1,令,-1=9,得,n,=99,故选B.,考点突破,2.数列,a,n,前,n,项和为,S,n,若,a,n,=,则,S,5,等于,(),A.1B.,C.,D.,答案,B,a,n,=,=,-,S,5,=,a,1,+,a,2,+,+,a,5,=1-,+,-,+,+,-,=,.,5/23,3.设数列,a,n,是首项为1等比数列,若,是等差数列,则,+,+,+,值等于,(),A.2 012B.2 013C.3 018D.3 019,答案,C设数列,a,n,公比为,q,由等差数列定义可得,+,=,则,+,=,解得,q,=1,则,a,n,=1,从而,+,=,故,+,+,+,=2 012,=3 018,.,6/23,4.数列,a,n,前,n,项和为,S,n,已知,S,n,=1-2+3-4+,+(-1),n,-1,n,则,S,17,=,.,答案,9,解析,S,17,=1-2+3-4+5-6+,+15-16+17=1+(-2+3)+(-4+5)+(-6+7)+,+(-14,+15)+(-16+17)=1+1+1+,+1=9.,5.若数列,a,n,通项公式为,a,n,=2,n,+2,n,-1,则它前,n,项和,S,n,=,.,答案,2,n,+1,-2+,n,2,解析,S,n,=(2,1,+1)+(2,2,+3)+(2,3,+5)+,+(2,n,+2,n,-1)=(2,1,+2,2,+,+2,n,)+1+3+5+,+(2,n,-1)=,+,=2,n,+1,-2+,n,2,.,7/23,考点一分组转化法求和,典例1,(北京,15,13分)已知,a,n,是等差数列,b,n,是等比数列,且,b,2,=,3,b,3,=9,a,1,=,b,1,a,14,=,b,4,.,(1)求,a,n,通项公式;,(2)设,c,n,=,a,n,+,b,n,求数列,c,n,前,n,项和.,解析,(1)等比数列,b,n,公比,q,=,=,=3,(1分),所以,b,1,=,=1,b,4,=,b,3,q,=27.,(3分),设等差数列,a,n,公差为,d,.,因为,a,1,=,b,1,=1,a,14,=,b,4,=27,所以1+13,d,=27,即,d,=2.,(5分),所以,a,n,=2,n,-1(,n,N,*,).,(6分),考点突破,8/23,(2)由(1)知,a,n,=2,n,-1,b,n,=3,n,-1,.,所以,c,n,=,a,n,+,b,n,=2,n,-1+3,n,-1,.,(8分),从而数列,c,n,前,n,项和,S,n,=1+3+,+(2,n,-1)+1+3+,+3,n,-1,=,+,=,n,2,+,.,(13分),9/23,规律总结,分组转化求和常见类型,(1)若,a,n,=,b,n,c,n,且,b,n,c,n,为等差或等比数列,可采取分组求和法求,a,n,前,n,项和.,(2)若,a,n,=,且数列,b,n,c,n,是等比数列或等差数列,可采取分,组求和法求和.,10/23,1-1,(烟台模拟),+,+,+,+,+,=,.,答案,-,+1,解析,因为,=,n,+,所以,+,+,+,+,+,=,+,+,+,+,+,=(1+2+3+,+,n,)+,=,+,=,-,+1.,11/23,考点二并项法求和,典例2,已知数列,a,n,前,n,项和为,S,n,=1-5+9-13+17-21+,+(-1),n,-1,(4,n,-3),则,S,15,+,S,22,-,S,31,值是,(),A.13B.-76C.46D.76,答案,B,解析,S,15,=1-5+9-13+,+(4,13-3)-(4,14-3)+(4,15-3)=7,(-4)+57=29,S,22,=1-5+9-13+,+(4,21-3)-(4,22-3)=11,(-4)=-44，,S,31,=1-5+9-13+,+(4,29-3)-(4,30-3)+(4,31-3)=15,(-4)+121=61，,S,15,+,S,22,-,S,31,=29-44-61=-76.,故选B,.,12/23,规律总结,并项求和解题思绪,并项求和常见有首末并项、隔项并项、分段并项、类周期并项,求解,时要注意观察通项结构特点,依据特点采取对应方法求解.,2-1,若数列,a,n,通项公式是,a,n,=(-1),n,(3,n,-2),则,a,1,+,a,2,+,+,a,10,=,(),A.15B.12C.-12D.-15,答案,A记,b,n,=3,n,-2,则数列,b,n,是以1为首项,3为公差等差数列,所,以,a,1,+,a,2,+,+,a,9,+,a,10,=(-,b,1,)+,b,2,+,+(-,b,9,)+,b,10,=(,b,2,-,b,1,)+(,b,4,-,b,3,)+,+(,b,10,-,b,9,)=5,3=,15.故选A.,13/23,考点三裂项相消法求和,典例3,(课标,17,12分),S,n,为数列,a,n,前,n,项和.已知,a,n,0,+2,a,n,=,4,S,n,+3.,(1)求,a,n,通项公式;,(2)设,b,n,=,求数列,b,n,前,n,项和.,解析,(1)由,+2,a,n,=4,S,n,+3,可知,+2,a,n,+1,=4,S,n,+1,+3.,可得,-,+2(,a,n,+1,-,a,n,)=4,a,n,+1,即,2(,a,n,+1,+,a,n,)=,-,=(,a,n,+1,+,a,n,)(,a,n,+1,-,a,n,).,由,a,n,0,可得,a,n,+1,-,a,n,=2.,又,+2,a,1,=4,a,1,+3,解得,a,1,=-1(舍去)或,a,1,=3.,所以,a,n,是首项为3,公差为2等差数列,通项公式为,a,n,=2,n,+1.,(6分),14/23,(2)由,a,n,=2,n,+1可知,b,n,=,=,=,.,设数列,b,n,前,n,项和为,T,n,则,T,n,=,b,1,+,b,2,+,+,b,n,=,=,.,(12分),15/23,易错警示,利用裂项相消法求和时,应注意抵消后并不一定只剩下第一项和最终一,项,也有可能前面剩两项,后面也剩两项.有些情况下,裂项时需要调整前,面系数,使裂开两项之差和系数之积与原通项相等.,16/23,3-1已知数列前四项是,则该数列前,n,项和,为,.,答案,-,解析,由题意可知通项,a,n,=,=,所以此数列前,n,项和,S,n,=,+,+,+,+,+,=,=,-,.,17/23,变式3-2,若本例(2)条件变为,b,1,=,n,2时,b,n,=(-1),n,-1,求数列,b,n,前,n,项和,T,n,.,解析,当,n,2时,b,n,=(-1),n,-1,=(-1),n,-1,当,n,=1时,b,1,=,也满足上式.,当,n,为偶数时,T,n,=,-,+,-,+,-,所以,T,n,=1-,=,.,18/23,当,n,为奇数时,T,n,=,-,+,-,-,+,所以,T,n,=1+,=,.,综上,T,n,=,19/23,考点四错位相减法求和,典例4,(山东,18,12分)已知数列,a,n,前,n,项和,S,n,=3,n,2,+8,n,b,n,是等,差数列,且,a,n,=,b,n,+,b,n,+1,.,(1)求数列,b,n,通项公式;,(2)令,c,n,=,求数列,c,n,前,n,项和,T,n,.,解析,(1)由题意知,当,n,2时,a,n,=,S,n,-,S,n,-1,=6,n,+5.,当,n,=1时,a,1,=,S,1,=11,所以,a,n,=6,n,+5.,设数列,b,n,公差为,d,.,由,即,可解得,b,1,=4,d,=3.所以,b,n,=3,n,+1.,20/23,(2)由(1)知,c,n,=,=3(,n,+1)2,n,+1,.,又,T,n,=,c,1,+,c,2,+,+,c,n,得,T,n,=3,2,2,2,+3,2,3,+,+(,n,+1),2,n,+1,2,T,n,=3,2,2,3,+3,2,4,+,+(,n,+1),2,n,+2,两式作差,得-,T,n,=3,2,2,2,+2,3,+2,4,+,+2,n,+1,-(,n,+1),2,n,+2,=3,=-3,n,2,n,+2,.,所以,T,n,=3,n,2,n,+2,.,21/23,方法技巧,(1)普通地,假如数列,a,n,是等差数列,b,n,是等比数列,求数列,a,n,b,n,前,n,项和时,可采取错位相减法求和,普通是和式两边同乘以等比数列,b,n,公比,然后作差求解;,(2)在写“,S,n,”与“,qS,n,”表示式时应尤其注意将两式“错项对齐”,方便下一步准确地写出“,S,n,-,qS,n,”表示式.,22/23,4-1,(广东广雅中学期中)已知,a,n,0,且,a,p,a,q,=2,p,+,q,(,p,q,N,*,)成立.,(1)求数列,a,n,通项公式;,(2)若,b,n,=,na,n,(,n,N,*,),求数列,b,n,前,n,项和,S,n,.,解析,(1)对于任意正整数,p,q,都有,a,p,a,q,=2,p,+,q,成立,令,p,=,q,=,n,可得,a,n,a,n,=2,2,n,因为,a,n,0,所以,a,n,=2,n,.,(2),b,n,=,na,n,=,n,2,n,S,n,=1,2,1,+2,2,2,+3,2,3,+,+(,n,-1)2,n,-1,+,n,2,n,2,S,n,=1,2,2,+2,2,3,+3,2,4,+,+(,n,-1)2,n,+,n,2,n,+1,-,得-,S,n,=2,1,+2,2,+2,3,+,+2,n,-,n,2,n,+1,即-,S,n,=,-,n,2,n,+1,整理可得,S,n,=(,n,-1),2,n,+1,+2.,23/23,