設(shè)等比數(shù)列{an}的公比為q,則其和Sn,S2n,S3n之間有以下關(guān)系:
Sn,S2n-Sn,S3n-S2n成等比數(shù)列,公比為q^n.
證明:先證明一個(gè)更一般的通項(xiàng)公式.在等比數(shù)列中,
an=a1q^(n-1)
am=a1q^(m-1)
兩式相除得an/am=q^(n-m),∴an=amq^(n-m).
S2n=a1+a2+...+an+a(n+1)+a(n+2)+...+a2n
=Sn+(a1q^n+a2q^n+...+anq^n)=Sn+(a1+a2+...+an)q^n=Sn+Snq^n
∴(S2n-Sn)/Sn=q^n.
同理,S3n=S2n+[a(2n+1)+a(2n+2)+...+a3n]
=S2n+[a(n+1)q^n+a(n+2)q^n+...+a2nq^n)
=S2n+[a(n+1)+a(n+2)+...+a2n]q^n=S2n+[S2n-Sn}q^n.