Question

若{an}为等比数列则Sn,S2n-Sn,S3n-S2n成等比数列

求证以上定理

Answer 1

解:设等比数列{an}的公比为q,
则Sn,S2n-Sn,S3n-S2n成等比数列,公比为q^n.

证明:先证明一个更一般的通项公式.在等比数列中,

an=a1q^(n-1)

am=a1q^(m-1)

两式相除得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.

∴(S3n-S2n)/(S2n-Sn)=q^n.

∴(S2n-Sn)/Sn=(S3n-S2n)/(S2n-Sn).即(S2n-Sn)^2=Sn(S3n-S2n).