请问sn s2n- sn s3n- s2n为什么是等差数列？ {an}为等差数列，则Sn,S2n-Sn,S3n-S2n也为...
an\u4e3a\u7b49\u5dee\u6570\u5217\uff0csn\uff0cs2n-sn\uff0cs3n-s2n\u4e3a\u4ec0\u4e48\u4e5f\u662f\u7b49\u5dee\u6570\u5217Sn=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\u65f6,
[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 \u603b\u5171n\u9879
=n²d,
\u6240\u4ee5\u4eceSn\u5f00\u59cb\u5c31\u662f\u7b49\u5dee.
Sn=a1+a2+a3+......+an;
S2n-Sn=(a1+a2+a3+......+an+......+a2n)-(a1+a2+a3+......+an)=an+1+an+2+......+a2n;
S3n-S2n=(a1+a2+a3+......+an+......+a2n+......+a3n)-(a1+a2+a3+......+an+......+a2n)=a2n+1+a2n+2+......+a3n;
\u516c\u5dee\uff1a
(S2n-Sn)-Sn=(an+1+an+2+......+a2n)-(a1+a2+a3+......+an)=(an+1-a1)+(an+2-a2)+......(a2n-an)=n*n*d;
(S3n-S2n)-(S2n-Sn)=(a2n+1+a2n+2+......+a3n)-(an+1+an+2+......+a2n)=(a2n+1-an+1)+(a2n+2-an+2)+......(a3n-a2n)=n*n*d;
\u5982\u679c{an}\u6210\u7b49\u6bd4\u65f6\u5019\uff0c\u4e0a\u5f0f\u597d\u50cf\u4e0d\u6210\u7b49\u6bd4\u6570\u5217\u5427\uff1f
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
扩展阅读：sn s2n-sn s3n-s2n成等差 ... www.sony.com.cn ... 等比数列sn s2n s3n ... an为等比数列sn s2n-sn ... sn s2n-sn的公差 ... 文学作者区sis0001邀请 ... 17173.com ... 新手s入门教程 ... s2n-sn成等差数列 ...
