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数列{An}中,A1=8,A4=2且满足An+2=2An+1-An,(n∈...

(1)∵an+2-2an+1+an=0(n∈N*)∴an+2-an+1=an+1-an∴{an}为等差数列,设其公差为d…(1分)又a1=8,a4=2,∴8+3d=2,∴a1=8,d=-2∴an=-2n+10 …(3分)(2)∵an=-2n+10,∴n≤5时,an≥0;n≥6时,an<0…(4分)∴n≥6时,Sn=|a1|+|a2|+…+|an|=a1+a2+…+a5-a...

解: a(n+2)=2a(n+1)-an a(n+2)-a(n+1)=a(n+1)-an 数列{an}是等差数列 a4=a1+3d a1=8,a4=2 d=(a4-a1)/3=(2-8)/3=-2 an=a1+(n-1)d =8+(-2)(n-1) =-2n+10 数列{an}的通项公式为an=-2n+10

an+2-2an+1+an=0 an+2-an+1=an+1-an 即数列an是等差数列 a4=a1+3d, 即2=8+3d,解得d=-2 an=a1+(n-1)d =8-2(n-1) =10-2n

解: (1) a(n+1)²=2an²+ana(n+1) a(n+1)²-ana(n+1)-2an²=0 数列各项均为正,an≠0,等式两边同除以an² [a(n+1)/an]²- a(n+1)/an -2=0 [a(n+1)/an +1][a(n+1)/an -2]=0 a(n+1)/an=-1(正项数列,比值为正,舍去)或a...

an+1-an=an+12+n+1-an2-n=2na+a+1当n≤4时,2na+a+1>0a>-12n+1≥-1/9当n≥8时,2na+a+1<0a<-12n+1≤-117,因此,-19<a<?117.答案:-19<a<?117.

(1)因为an+12=2an2+anan+1,即(an+1+an)(2an-an+1)=0,又an>0,所以有2an-an+1=0,所以2an=an+1,所以数列{an}是公比为2的等比数列.由a2+a4=2a3+4得2a1+8a1=8a1+4,解得a1=2,故an=2n(n∈N*)(2)nan=n?2n,Sn=2+2?22+3?23+…+(n-1)?2...

(Ⅰ)因为a2n+1=2a2n+anan+1,所以(an+1+an)(2an-an+1)=0,因为an>0,?所以有2an-an+1=0,即2an=an+1,所以数列{an}是公比为2的等比数列,?由a2+a4=2a3+4得2a1+8a1=8a1+4,解得a1=2.从而数列{an}的通项公式为an=2n.…(6分)(II)bn=na...

解(Ⅰ)a2=5,a3=7,a4=9,猜想an=2n+1.…(4分)(Ⅱ)Sn=n(3+2n+1)2=n2+2n,…(6分)使得2n>Sn成立的最小正整数n=6.…(7分)下面给出证明:n≥6(n∈N*)时都有2n>n2+2n.①n=6时,26>62+2×6,即64>48成立;…(8分)②假设n=k(k≥6,k∈N*)时...

检查一下抄错题了吧,a1应该等于3吧?

1.因为数列{an}的前n项和Sn=2an-2^n....(1)所以S(n+1)=2a(n+1)-2^(n+1)....(2)(2)-(1)得a(n+1)=2a(n+1)-2an-2^n所以a(n+1)-2an=2^n所以(a(n+2)-2a(n+1))/(a(n+1)-2an)=2^(n+1)/2^n=2所以数列{a(n+1)-2an}是等比数列2.因为a(n+1)-2an=2^n两边同时...

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