2025优化设计一轮课时规范练47　数列中的奇、偶项问题

课时规范练47数列中的奇、偶项问题1.(2024·云南昆明模拟)在数列{an}中,a1=5,当n≥2时,an=2an-1+2n-1.(1)求证:数列{an(2)设bn=(-1)nlog2(an-1n+1),数列{bn}的前n项和为S2.已知数列{an}满足a1=1,an+1=an+1,n为奇数,2an从①bn=a2n-1+2,②bn=a2n+1-a2n-1这两个条件中任选一个填在横线上,并完成下面问题.(1)写出b1,b2,并求数列{bn}的通项公式;(2)求数列{an}的前n项和Sn.3.(2024·河南开封模拟)已知数列{an}的前n项和Sn满足2Sn=(n+1)an,且a1=1.(1)求{an}的通项公式;(2)若bn=(-1)nan2,求数列{bn}的前n项和T4.已知数列{an}满足a1=1,an+1=a(1)记bn=a2n,写出b1,b2,并求数列{bn}的通项公式;(2)求{an}的前n项和Tn.

课时规范练47数列中的奇、偶项问题1.(1)证明因为an=2an-1+2n-1(n≥2),所以(an-1)-2(an-1-1)=2n,两边同除以2n,得an-12n所以{an-12n}是以a1(2)解由(1)知,an-12n=2+(n-1)=n+1,整理得an=2n(n+1)+1,则bn=(-1)nlog2(an-1n+1)=(-1)nlog当n为偶数时,Sn=(-1+2)+(-3+4)+…+[-(n-1)+n]=n2当n为奇数时,Sn=Sn+1-bn+1=n+12-(n+1)所以Sn=-2.解(1)若选①,bn+1=a2n+1+2=2a2n+2=2(a2n-1+1)+2=2(a2n-1+2)=2bn,且b1=a1+2=3,故数列{bn}是首项为3,公比为2的等比数列,bn=3·2n-1,故b2=6.若选②,a2n+1=2a2n=2(a2n-1+1)=2a2n-1+2,所以a2n+1+2=2(a2n-1+2),且a1+2=3,故数列{a2n-1+2}是以3为首项,2为公比的等比数列,所以a2n-1+2=3·2n-1,故a2n-1=3·2n-1-2,所以bn=a2n+1-a2n-1=(3·2n-2)-(3·2n-1-2)=3·2n-1,故b1=3,b2=6.(2)由(1)可知a2n-1=3·2n-1-2,则a2n=a2n-1+1=3·2n-1-1,所以a2n-1+a2n=3·2n-3.当n为偶数时,Sn=(a1+a2)+(a3+a4)+…+(an-1+an)=(3·21+3·22+…+3·2n2)-3·n2=6(1-2当n为奇数时,Sn=(a1+a2)+(a3+a4)+…+(an-2+an-1)+an=3·2n-1+22−3(n-综上所述,Sn=33.解(1)因为2Sn=(n+1)an,当n≥2时,2Sn-1=nan-1,所以2Sn-2Sn-1=(n+1)an-nan-1,即2an=(n+1)an-nan-1,所以(n-1)an=nan-1,所以ann=an-1n-1,即{ann}是常数列.又a(2)bn=(-1)nan2=(-1)n×n当n为偶数时,Tn=-12+22-32+42-…-(n-1)2+n2=(22-12)+(42-32)+…+[n2-(n-1)2]=(2-1)×(2+1)+(4-3)×(4+3)+…+[n-(n-1)][n+(n-1)]=2+1+4+3+…+n+(n-1)=(1+当n为奇数时,Tn=-12+22-32+42-…+(n-1)2-n2=(22-12)+(42-32)+…+[(n-1)2-(n-2)2]-n2=(2-1)×(2+1)+(4-3)×(4+3)+…+[(n-1)-(n-2)][(n-1)+(n-2)]-n2=2+1+4+3+…+(n-2)+(n-1)-n2=[1+(n-综上可得,Tn=n4.解(1)因为数列{an}满足a1=1,an+1=a所以,b1=a2=a1+1=2,b2=a4=a3+1=a2+2+1=5.bn+1=a2n+2=a2n+1+1=a2n+2+1=a2n+3=bn+3,即bn+1-bn=3,所以数列{bn}是公差为3,首项为2的等差数列,因此bn=2+3(n-1)=3n-1.(2)当n为偶数时,设n=2k(k∈N*),则a2k=a2k-1+1,k=n2,所以a2k+a2k-1=2a2k-1=2bk-1=2(3k-1)-1=6k-此时,Tn=T2k=(a1+a2)+(a3+a4)+…+(a2k-1+a2k)=(3+6k-3)k2=3k2