在数列an中,a1=1,前n项和Sn=n 2 3an,求数列
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由于:a(n+1)=2an+3则有:[a(n+1)+3]=2(an+3)[a(n+1)+3]/(an+3)=2则:{an+3}为公比为2的等比数列则:an+3=(a1+3)*2^(n-1)=4*2^(
(1)数列{an}中,a1=1,前n项和Sn=n+23an,可知S2=43a2,得3(a1+a2)=4a2,解得a2=3a1=3,由S3=53a3,得3(a1+a2+a3)=5a3,解得a3=32(a
(1)an=sn-s(n-1)就有sn-s(n-1)+2sn*s(n-1)=0两边同除以sn*s(n-1)得1/sn-1/s(n-1)=2{1/sn}是等差数列1/sn=1/s1+(n-1)d=2n-
设:(An+1)+p(n+1)+q=4[An+pn+q]解得p=-1,q=0即An+1=4An-3n+1等价于(An+1)-(n+1)=4(An-n)若设Bn=An-n则Bn+1=4Bn则Bn=B1*
a(n+1)=a(n)+2说明这是一个等差数列首项a(1)=-11,公差为2a(n)=a(1)+(n-1)×2=-11+2(n-1)=2n-13所以Sn=[a(1)+a(n)]×n/2=(n-12)n
S[1]=a[1]=1/2(a[1]+1/a[1]),于是:a[1]=1=√1-√0S[2]=a[2]+1=1/2(a[2]+1/a[2]),于是:a[2]=√2-1,S[2]=√2S[3]=a[3]
当公比为1时,Sn=n,数列{Sn+12}为数列{n+12}为公差为1的等差数列,不满足题意;当公比不为1时,Sn=1−qn1−q,∴Sn+12=1−qn1−q+12,Sn+1+12=1−qn+11−
1/a(n+1)=an+2/2an=1/2+1/an所以,{1/an}是公差为1/2的等差数列1/an=1/a1+(n-1)*1/2=(n+1)/2an=2/(n+1)a(n+1)=2/(n+3)an
s(n+1)=(n+1)^2Xa(n+1)=sn+a(n+1)=n^2Xan+a(n+1)就有下列等式a(n+1)/an=n/(n+2)我们可以得出a(n+1)=a(n+1)/anXan/a(n-1)
a(n+1)=3a(n)/[a(n)+3],若a(n+1)=0,则,a(n)=0,...,a(1)=0,与a(1)=1/2矛盾.因此,a(n)不为0.1/a(n+1)=(1/3)[a(n)+3]/a(
a1=Sna1=S1=2an=Sn-Sn-1=n²+n-(n-1)²-(n-1)=2nan=2na2=2*2=4a3=2*3=6a1=2,a2=4,a3a=6an=2n
在数列{an}中,a1=1,a(n+1)=2an+2^n,求数列的前n项和a(n+1)=2an+2^n同除以2^na(n+1)/2^n=2an/2^n+1a(n+1)/2^n-an/2^(n-1)=1
a(n+1)=(1+1/n)an+(n+1)/2^n,故a(n+1)/(n+1)=an/n+1/2^n,用累加法得an/n-a1/1=1-1/2^(n-1)即an/n=1-1/2^(n-1)+a1故a
证:a(n+1)=2an/(an+1)1/a(n+1)=(an+1)/(2an)=(1/2)(1/an)+1/21/a(n+1)-1=(1/2)(1/an)-1/2=(1/2)(1/an-1)[1/a
将a[n+1]=S[n+1]-S[n]代人得到:S[n]=4(S[n+1]-S[n])+14S[n+1]=5S[n]-14(S[n+1]-1)=5(S[n]-1)(S[n+1]-1)/(S[n]-1)
s(n)/n=(2n-1)a(n),s(n)=n(2n-1)a(n),a(n+1)=s(n+1)-s(n)=(n+1)(2n+1)a(n+1)-n(2n-1)a(n),[2n^2+3n]a(n+1)=
an+1=(1/2)an+1还是an+1=1/(2an)+1如果是前者,用配方法an+1-2=(an-2)/2{an-2}是等比数列an-2=(a1-1)/2^(n-1)an=4*2^(-n)+2Sn
题为:在数列{a[n]}中,a[1]=2,a[n+1]=a[n]+cn(c是常数),且a[1]、a[2]、a[3]成等比数列,求数列{(a[n]-c)/(n.c^n)}的前n项之和T[n].其中[&n
an+a(n+1)=6/5^(n+1)=(5+1)/5^(n+1)=1/5^n+1/5^(n+1)a(n+1)-1/5^(n+1)=-(an-1/5^n)a1-1/5^1=1/5-1/5=0an-1/
看到这类题目,求的项数很大的,通常是可以累积错位抵消或者循环.是选择或者填空,先把a1带进等式求出几个,a2=-1/2,a3=2/3,a4=3,求到a4就可以看出来了,每3项就是一个循环.如果是问答题