n=1有s1=a1=3/2(a1-1) ==> a1=3
n>1有an=Sn-Sn-1 = 3/2 (an - 1) - 3/2 (an-1 - 1) ==>an=3*an-1 公比3的等比数列
an = 3^n;
bn=an*log an = 3^n * n* log3
利用错位相减可以求出bn的和
已知数列{an}的前n项和为Sn,且满足Sn=3/2(an-1)(n∈正整数) 若bn=anlogan,求数列bn的前n
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