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
人气：309 ℃ 时间：2019-08-20 00:38:56
解答
令S1=a1=tS2=a1+a2=2a1+2=2t+2S3=a1+a2+a3=3a1+6=3t+62√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)16t...
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