an= n*2^n = n* 2^(n+1) - n* 2^n = n* 2^(n+1) - ( n-1) * 2^n - 2^n设 bn=2^n的前n项的和为TnTn =2 (2^n-1)=2^(n+1) -2 Sn= n* 2^(n+1) -0 - Tn = n*2^(n+!) - 2^(n+1) +2Sn = (n-1)*2^(n+1) + 2
