sn2n- sn3n- s2n是不是等差数列

Question

如题所述

Answer 1

若an为等差数列
则Sn=na1+n(n-1)d/2,
S2n=2na1+2n(2n-1)d/2,
S2n-Sn=na1+n(3n-1)d/2,
(S2n-Sn)-Sn=n²d,
k>1时,
[Skn -S(k-1)n]-[S(k-1)n -S(k-2)n]
={a[(k-1)n+1] +a[(k-1)n+2]+...+a[kn] } - {a[(k-2)n+1] +a[(k-2)n+2]+...+a[(k-1)n] }
={a[(k-1)n+1] -a[(k-2)n+1] }+ {a[(k-1)n+2] -a[(k-2)n+2]}+...+{a[kn] -a[(k-1)n] }
=nd+nd+...+nd 总共n项
=n²d 所以sn s2n-sn s3n-s2n 也是等差数列 公差为n*2d
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). 所以sn s2n-sn s3n-s2n 是等比数列 打字辛苦 望采纳