等比数列{an}的首项是a1,公比是q,前n项之和为Sn,所有项的和为S,则lim(S1+S2+...+Sn-nS)=_______数学解答

等比数列{an}的首项是a1,公比是q,前n项之和为Sn,所有项的和为S,则lim(S1+S2+...+Sn-nS)=______.

人气：447 ℃　时间：2019-12-15 05:27:40

解答

S1=a1(1-q)/(1-q),S2=a1(1-q^2)/(1-q),...,Sn=a1(1-q^n)/(1-q).S1+S2+...+Sn=[a1/(1-q)]*[1-q+1-q^2+...+1-q^n]=[a1/(1-q)]*[n-q(1-q^n)/(1-q)]=na1/(1-q)-a1q(1-q^n)/(1-q)^2,nS=na1/(1-q),(S1+S2+...+Sn)-nS=-a1q(...