设等比数列 {an} 的公比为q,前n项和为Sn,若S(n+1),Sn,S(n+2)成等差数列,则q=
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设等比数列 {an} 的公比为q,前n项和为Sn,若S(n+1),Sn,S(n+2)成等差数列,则q=
a(n)=aq^(n-1),n=1,2,...
若q=1.则s(n)=na,n=1,2,...
s(n+1)+s(n+2)-2s(n)=(n+1)a+(n+2)a-2na=3a不等于0,矛盾.
因此,q不为1.此时,
s(n)=a[q^n-1]/(q-1),n=1,2,...
0=s(n+1)+s(n+2)-2s(n)=a[q^(n+1)-1]/(q-1)+a[q^(n+2)-1]/(q-1)-2a[q^n-1]/(q-1),
0=q^(n+1)-1+q^(n+2)-1-2[q^n-1]=q^n[q+q^2-2]=q^n(q-1)(q+2),
q=-2.
若q=1.则s(n)=na,n=1,2,...
s(n+1)+s(n+2)-2s(n)=(n+1)a+(n+2)a-2na=3a不等于0,矛盾.
因此,q不为1.此时,
s(n)=a[q^n-1]/(q-1),n=1,2,...
0=s(n+1)+s(n+2)-2s(n)=a[q^(n+1)-1]/(q-1)+a[q^(n+2)-1]/(q-1)-2a[q^n-1]/(q-1),
0=q^(n+1)-1+q^(n+2)-1-2[q^n-1]=q^n[q+q^2-2]=q^n(q-1)(q+2),
q=-2.
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