Question

高中数学题 数列的

等比数列{an}的首项a1=1，公比为q且满足q的绝对值<1，前n项之和为Sn，各项之和为S。求：lim（S1+S2+。。。+Sn-nS）

Answer 1

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

祝您学习愉快

Answer 2

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

Answer 3

S＝1/(1-q)
Sn=(1-q^n)/(1-q)
Sn-S=q^n/(q-1)
lim(S1+S2+。。。+Sn-nS)
=lim[(S1-S)+(S2-S)+……+(Sn-S)]
=1/(q-1)·lim(q+q^2+……+q^n)
=1/(q-1)·lim[q(1-q^n)/(1-q)]
=-q/(1-q)^2·lim(1-q^n)
=-q/(1-q)^2