已知等比数列{an}满足2a1+a3=3a2,且a3+2是a2,a4的等差中项

因为an=2^n,所以 log2 1/an（2为角标）= -n所以 bn=2^n-nSn=2-1+2^2-2+2^3-3+...+2^n-n =(2+2^2+2^3+...+2^n)-(1+2+3+...+n)= 2^(n+1)-2-(1+n)*n/2Sn-2^(n+1)+47=2^(n+1)-2-(1+n)*n/2-2^(n+1)+47=45-(1+n)*n/290(n+1...