设数列(an)是首项为a1(a＞0),公差为2的等差数列,其前n项和为Sn,且√s1,√s2,√s3

来源：学生作业帮编辑：搜狗做题网作业帮分类：数学作业时间：2024/05/02 13:57:06

设数列(an)是首项为a1(a＞0),公差为2的等差数列,其前n项和为Sn,且√s1,√s2,√s3
成等差数列,(1)求(an)的通项公式(2)记bn=(an)/(2^n)的前n项和为Tn,求Tn

令S1=a1=t
S2=a1+a2=2a1+2=2t+2
S3=a1+a2+a3=3a1+6=3t+6
2√S2=√S1+√S3,
2√(2t+2)=√t+√(3t+6),
4(2t+2)=t+3t+6+2√[t(3t+6)]
8t+8=4t+6+2√(3t²+6t)
4t+2=2√(3t²+6t)
16t²+16t+4=4(3t²+6t)
16t²+16t+4=12t²+24t
4t²-8t+4=0
t²-2t+1=0
(t-1)²=0
t=1
即a1=1
an=a1+(n-1)d=1+2(n-1)=2n-1
∴an=2n-1
bn=(an)/(2^n)=(2n-1)/(2^n)
Tn=b1+b2+b3+...+bn
2Tn=2b1+2b2+2b3+...+2bn
2Tn-Tn
=(2b1+2b2+2b3+...+2bn)-(b1+b2+b3+...+bn)
=2b1+(2b2-b1)+(2b3-b2)+...+(2bn-b[n-1])-bn
=1+(3/2-1/2)+(5/4-3/4)+...+[(2n-1)/2^(n-1)-(2n-3)/2^(n-1)] - (2n-1)/2^n
=1+1+[1/2+1/4+...+1/2^(n-2)] - (2n-1)/2^n
=2+ [1-1/2^(n-2)] - (2n-1)/2^n
=3- (2n+3)/2^n
∴Tn=3- (2n+3)/2^n

设数列(an)是首项为a1(a＞0),公差为2的等差数列,其前n项和为Sn,且√s1,√s2,√s3 设数列{an}是首项为a1(a1>0),公差为2的等差数列,前n项和为Sn,且根号S1,根号S2,根号S3成等差数列, 设数列an是首项为a1(a1>0),公差为2的等差数列,其前n项和为Sn,且根号S1,根号S2.根号S3成等差数列.求a 设数列{an}是首项为a1(a1>0),公差为2的等差数列,其前n项和为Sn,且根号S1,S2,S3成等差数列.求数列{ 设数列{An}是首项为a1,a1>0,公差为2地等差数列,其前N项和为Sn,且根号S1,根号S2,根号S3成等差数列.求 #高考提分#设数列{an}是首项为a1(a1>0),公差为2的等差数列,前n项和为Sn,且根号S1,根号S2,根号S3成数列an是首项为3公差为2的等差数列其前n项和为Sn求An=1/S1+1/S2+1/S3+...+1/Sn 已知等差数列{an的公差为2,前n项和为Sn,且S1,S2,S3成等比数列.(1)求数列{an的通项公式已知等差数列｛an｝的公差为2,前n项和为sn,且s1,s2,s3,s4成等比数列设数列{an}的前n项和为Sn,且an不等于0,S1,S2,S3 Sn成等比数列,试问a1,a2,a2是等比数列吗等比数列{an}中,前n项和为sn,已知S1,S3,S2成等差数列.问：若a1-a3=3,求数列S1、S3、S2的公差d 已知等差数列{an}的公差d＞0，设{an}的前n项和为Sn，a1=1，S2•S3=36．