无穷等比数列{an}的前n项和Sn=a-(1/3)^n,则无穷等比数列求各项和与a值

Question

无穷等比数列{an}的前n项和Sn=a-(1/3)^n,则无穷等比数列 求各项和与a值

Answer 1

解 由无穷等比数列{an}的前n项和Sn=a-(1/3)^n 得 S1=a-1/3=a1
a2=S2-S1=a-(1/3)^2-(a-1/3)=1/3-1/9=2/9
a3=S3-S2=a-(1/3)^3-(a-(1/3)^2)=1/9-1/27=2/27
所以公比q=a3/a2=(2/27):2/9=1/3
所以a1=a2/1/3=3*2/9=2/3=a-1/3
a=1
Sn=(a1-anq)/(1-q)=2/3-2/3(1/3)^n/(1-1/3)=1-(1/3)^n