www.rtmj.net > 已知数列{An}是等差数列,其前n项和为Sn,若A1A2A3...

已知数列{An}是等差数列,其前n项和为Sn,若A1A2A3...

设等差数列的首项为,公差为d∵3 S1S3 +15 S3S5 +5 S5S1 =3 5 整理可得,15S5+75S1+25S3=3S1S3S5利用等差数列的前n项和公式可得,15(5a1+10d)+75a1+25(3a1+3d)=3a1(3a1+3d)(5a1+10d)即225(a1+d)=45a1(a1+d)(a1+2d)=a1a2a3*45=45*15解可得,a2=3故答案为:3

你确定上面的答案是正确的?你说的没错,我认为是这样的:我们可以知道a1=3,d=2,而Sp+q=3(p+q)+(p+q)(p+q-1)=2(p+q)+(p+q)^2(S2p+S2q)/2=2(p+q)+2p^2+2q^2由均值不等式,2p^2+2q^2>(p+q)^2(S2p+S2q)/2>Sp+q不知你怎么认为?

a2=a1+d,a3=a1+2d,a1=a1,a4=a1+3d因为a2*a3=28a1+a4=11所以(a1+d)(a1+2d)=28a1+(a1+3d)=11解得a1=1d=3所以an=1+(n-1)*3=3n-2

D 如图: 如果你认可我的回答,请点击“采纳答案”,祝学习进步! 如不明白请追问,手机提问的朋友在客户端右上角评价点【评价】,谢谢!

(1)设等比数列{an}的公比为q,由题意得,a1+a1q2=5a1q+a1q3=10,解得a1=1,q=2,∴an=2n?1,(2)由(1)得,bn=1+log4an=1+log2n?14=n+12,∴1bnbn+1=4(n+1)(n+2)=4(1n+1?1n+2),设数列{1bnbn+1}的前n项和为Tn,∴Tn=4[(12?13)+(13?14)+(14?15)+…+(1n+1?1n+2)]=4(12?1n+2)=2nn+2.

等差数列通项公式:an=a1+(n-1)d 等差数求和公式:Sn=na1+n(n-1)d/2 可以参考下这个网站的总结:http://baike.baidu.com/view/62268.html?wtp=tt解:根据等差数列的求和公式,得:S1=a1 (1) S3=3a1+3d (2) S5=5a1+10d (3)

S5=(a1+a5)*5/2 = 5a1+10d S10-S5=(a1+a10)*10/2-(a1+a5)*5/2 = 5(a1+a10) - 5(a1+a5) /2= 5a1/2 + 5a10 - 5a5/2= 5a10+35dS15-S10=5a1+60d 所以S5+(S15-S10)=2(S10-S5)

已知数列{an}是等差数列,其前n项和为sn,a3=7,s4=24 ,sn=na1+n(n-1)d,4a1+4(4-1)d=24, a1+2d=7,d=-1,a1=9,设p、q是正整数,且p≠q,s(p+q)=9(p+q)-(p+q)(p+q-1)=10(p+q)-(p+q), (s2p+s2q)/2 =[18p-2p(2p-1)+18q-2q(2q-1)]/2=10(p+q)-2(p+q),p+q>2pq, 2(p+q)>(p+q),-2(p+q)1/2(s2p+s2q) .

an =a1+(n-1)dSn =a1+a2++ana3=7a1+2d=7 (1)S4=24(2a1+3d).2 = 242a1+3d =12 (2)2(1)-(2)d=2a1=3an = 3+2(n-1)=2n+1bn = 2^an =2^(2n+1)Tn = b1+b2++bn = (8/3)( 2^(2n) -1)

S4=2(a2+a3)=2a2+14=24.a2=5.所以d=2.An=2n+1

网站地图

All rights reserved Powered by www.rtmj.net

copyright ©right 2010-2021。
www.rtmj.net内容来自网络,如有侵犯请联系客服。zhit325@qq.com