设等比数列{an}的前n项和sn,若sn=3^na+b,且a≠0,a,b为常数,则a+b=在数列{an}中,若a1=2,anan+1+an+1+1=0,则s2010=
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![设等比数列{an}的前n项和sn,若sn=3^na+b,且a≠0,a,b为常数,则a+b=在数列{an}中,若a1=2,anan+1+an+1+1=0,则s2010=](/uploads/image/z/10419102-54-2.jpg?t=%E8%AE%BE%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8Csn%2C%E8%8B%A5sn%3D3%5Ena%2Bb%2C%E4%B8%94a%E2%89%A00%2Ca%2Cb%E4%B8%BA%E5%B8%B8%E6%95%B0%2C%E5%88%99a%2Bb%3D%E5%9C%A8%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%EF%BC%8C%E8%8B%A5a1%3D2%EF%BC%8Canan%2B1%2Ban%2B1%2B1%3D0%EF%BC%8C%E5%88%99s2010%3D)
设等比数列{an}的前n项和sn,若sn=3^na+b,且a≠0,a,b为常数,则a+b=在数列{an}中,若a1=2,anan+1+an+1+1=0,则s2010=
设等比数列{an}的前n项和sn,若sn=3^na+b,且a≠0,a,b为常数,则a+b=
在数列{an}中,若a1=2,anan+1+an+1+1=0,则s2010=
设等比数列{an}的前n项和sn,若sn=3^na+b,且a≠0,a,b为常数,则a+b=在数列{an}中,若a1=2,anan+1+an+1+1=0,则s2010=
a1=S1=a+b
n>1时 an=Sn-S(n-1)=a*3^n+b-[a*3^(n-1)+b]=2a*3^(n-1)
a2=6a 等比q=3
故有3a1=a2 3(a+b)=6a a=b
前N项和为a1(3^n-1)/2=a*3^n-a
得a=b=0
故a+b=0
答案是0 ,
显然公比不为1,所以Sn=a1(1-q~n)/(1-q)
=a1/(1-q)-a1q~n/(1-q)
a=a1/(1-q) b= -a1/(1-q) 所以a+b=0
Sn=a1(1-q^n)/(1-q)=3^na+b
a1/(1-q)-[a1/(1-q)]q^n=3^na+b
两边对比可知:q=3,a=-[a1/(1-q)]=a1/2 b=a1/(1-q)=-a1/2
a+b=0