David_King98 幼苗
共回答了16个问题采纳率:93.8% 举报
∵Sn=a1(1−qn)1−q,q=2∴S4=a1(1−24)1−2=15a1=15,则a1=1∴S8=1−281−2=255故答案为:255
点评:本题考点: 等比数列的前n项和. 考点点评: 此题考查学生灵活运用等比数列的前n项和公式化简求值,是一道基础题.
1年前
回答问题
(2008•东城区二模)已知数列{an}为等差数列.
1年前1个回答
(2008•东城区二模)已知{an}是等比数列,a2=6,a3=12,则数列{an}的前n项的和S5=______.
(2008•西城区一模)设等差数列{an}的各项均为整数,其公差d≠0,a5=6.
(2008•西城区一模)数列{an}的通项公式为an=n+2n(n=1,2,3,…),则{an}的前n项和Sn=n(n+
(2010•东城区一模)已知数列{an},{bn},其中a1=12,数列{an}的前n项和Sn=n2an(n≥1),数列
(2013•东城区一模)已知数列{an}中,a1=2,an+1-2an=0,bn=log2an,那么数列{bn}的前10
(2007•东城区一模)已知{an}是首项为1,公比为q的等比数列,Pn=a1+a2C1n+a3C2n+…+an+1Cn
(2010•东城区二模)已知数列{an}中,Sn是其前n项和,若a1=1,a2=2,anan+1an+2=an+an+1
(2010•东城区二模)已知数列{an}的前n项和为Sn,a1=1,Sn+1=4an+1,设bn=an+1-2an.
(2007•东城区二模)已知等差数列{an}的公差为2,若a2,a4,a5成等比数列,则a2=______.
(2007•东城区一模)已知f(x)=(x-1)2,g(x)=10(x-1),数列{an}满足a1=2,(an+1-an
(2009•东城区一模)已知递增的等比数列{an}满足a2+a3+a4=28,且a3+2是a2,a4的等差中项.
(2013•东城区模拟)已知数列{an}是等差数列,a2=6,a5=18;数列{bn}的前n项和是Tn,且Tn+[1/2
(2010•东城区二模)已知数列{an}中,a1=b(b>0),an+1=−1an+1(n∈N*),能使an=b的n可以
(2011•东城区模拟)已知数列{an}为等差数列,a3=5,a7=13,数列{bn}的前n项和为Sn,且有Sn=2bn
(2013•东城区模拟)已知数列{an}是等差数列,a3=10,a6=22,数列{bn}的前n项和是Tn,且Tn+13b
(2006•东城区二模)已知数列{an}是等差数列,a2=6,a5=18,数列{bn}的前n项和是Tn,且Tn+12bn
(2003•东城区二模)已知数列{an}的各项均为正整数,且满足an+1=an2-2nan+2,(n∈N),又a5=11
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