这个数列怎么求和等比数列an=2^n,求S1+2S2+...+nSn

∵an=2^n
∴a1=2 q=2
于是Sn=2(1-2ⁿ)/(1-2)=2(2ⁿ-1)=2^(n+1)-2
S1+2S2+...+nSn
=2²-2+2*(2³-2)+3*(2⁴-2)+……+n[2^(n+1)-2]
=2²+2*2³+3*2⁴+……+n*2^(n+1)-(2+2*2+3*2+n*2)
=[2²+2*2³+3*2⁴+……+n*2^(n+1) ]-2(1+2+3+…………+n)
设Tn=2²+2*2³+3*2⁴+……+n*2^(n+1)
则2Tn= 2³+2*2⁴+3*2^5+……+n*2^(n+2)
相减得
Tn=-2²-2³-2⁴-……-2^(n+1)+n*2^(n+2)
=-[2²+2³+2⁴+……+2^(n+1)]+n*2^(n+2)
=-2²(1-2ⁿ)/(1-2)+n*2^(n+2)
=4-2^(n+2)+n*2^(n+2)
=(n-1)*2^(n+2) +4
Qn=2(1+2+3+…………+n)=2*(n+1)n/2=n²+n
∴S1+2S2+...+nSn
=Tn-Qn
=(n-1)*2^(n+2) +4-(n²+n)
=(n-1)*2^(n+2) -n^2-n+4

用错位相减法，具体运算挺复杂的

∵an=2^n
∴Sn＝2^(n＋1) －2
设Bn＝nSn＝n·2^(n＋1)－2n
则Tn＝S1+2S2+...+nSn
＝1·2²＋2·2³＋…＋n·2^(n＋1)－2×(1＋n)·n／2
＝1·2²＋2·2³＋…＋n·2^(n＋1) ＋ 0 ...