《高考总复习优化设计二轮复习》素养提升微专题(四)-数列解答题中的奇、偶项问题.pptx

高考总,复习优化设计,GAO KAO ZONG FU XI YOU HUA SHE JI,素养提升微专题,(,四,),数列解答题中,的,奇,、偶项问题,第一编,2022,规律方法,数列中的奇、偶项问题是对一个数列分成两个新数列进行单独研究,利用新数列的特征,(,等差、等比数列或其他特征,),求解原数列,.,数列中奇、偶项问题的常见题型有,:,(1),数列中连续两项的和或积的问题,(,a,n,+a,n+,1,=f,(,n,),或,a,n,a,n+,1,=f,(,n,);,(2),通项中含有,(,-,1),n,的类型,;,(3),a,2,n,a,2,n-,1,的类型,;,(4),已知明确的奇、偶项问题,.,考查角度,角度一,通项中含有,(,-,1),n,的数列求和,例,1(,2021,山东日照二模,),已知,正项,数列,a,n,其前,n,项和为,S,n,a,n,=,1,-,2,S,n,.,(1),求数列,a,n,的通项公式,;,角度二,奇、偶项通项不同的数列,求和,(1),求,a,2,a,3,及通项公式,a,n,;,(2),记,b,n,=a,n,+a,n+,1,求数列,2,n-,1,b,n,的前,2,n,项的和,T,2,n,.,所以,a,1,=S,1,=,1,a,2,=S,2,-S,1,=,4,-,1,=,3,a,3,=S,3,-S,2,=,3,-,4,=-,1,.,当,n,3,且,n为奇数时,a,n,=S,n,-S,n-,1,=n-,(,n-,1),2,=-n,2,+,3,n-,1,又,a,1,=,1,符合上式,所以当,n,为奇数时,a,n,=-n,2,+,3,n-,1,.,当,n,为偶数时,a,n,=S,n,-S,n-,1,=n,2,-,(,n-,1),=n,2,-n+,1,.,(2),由,(1),知,当,n,为奇数时,b,n,=a,n,+a,n+,1,=,4,n,当,n,为偶数时,b,n,=a,n,+a,n+,1,=,2,.,所以,T,2,n,=,1,2,2,+,3,2,4,+,+,(2,n-,1),2,2,n,+,2,2,+,2,4,+,+,2,2,n,=,2,22,+,4,2,4,+,+,2,n,2,2,n,所以,4,T,2,n,=,2,2,4,+,4,2,6,+,+,2,n,2,2,n+,2,对点演练,1,.,(2021,山东济南模拟,),已知数列,a,n,的前,n,项和为,S,n,且,2,S,n,=,3,a,n,-,3,.,(1),求数列,a,n,的通项公式,;,(2),设,b,n,=,2log,3,a,n,+,(,-,1),n,n,求数列,b,n,的前,n,项和,T,n,.,解,(1),当,n=,1,时,2,S,1,=,2,a,1,=,3,a,1,-,3,所以,a,1,=,3,.,因为,2,S,n,=,3,a,n,-,3,所以当,n,2,时,2,S,n-,1,=,3,a,n-,1,-,3,所以,2,S,n,-,2,S,n-,1,=,2,a,n,=,3,a,n,-,3,a,n-,1,即,=,3(,n,2),.,所以数列,a,n,是首项为,3,公比为,3,的等比数列,所以,a,n,=,3,n,.,2,.,(2021,浙江杭州二中月考,),设等差数列,a,n,的前,n,项和为,S,n,且,a,2,=,8,S,4,=,40,.,数列,b,n,的前,n,项和为,T,n,且,T,n,-,2,b,n,+,3,=,0,.,(1),求数列,a,n,b,n,的通项公式,;,因为,T,n,-,2,b,n,+,3,=,0,所以当,n=,1,时,b,1,-,2,b,1,+,3,=,0,即,b,1,=,3;,当,n,2,时,T,n-,1,-,2,b,n-,1,+,3,=,0,所以,T,n,-T,n-,1,-,2,b,n,+,2,b,n-,1,=-b,n,+,2,b,n-,1,=,0,即,b,n,=,2,b,n-,1,(,n,2),.,所以数列,b,n,是首项为,3,公比为,2,的等比数列,所以,b,n,=,3,2,n-,1,.,又,P,1,=c,1,=,4,符合上式,所以当,n,为奇数时,P,n,=,2,n,+n,2,+,2,n-,1,.,本 课 结 束,