在等差数列中。sn s2n-sn s3n-s2n怎样证明公差为n方d会的详细写下过程 已知等差数列{an}的前n项和为Sn，请证明Sn，S2n-S...

\u7ed9\u6211\u89e3\u91ca\u4e00\u4e0b\u8fd9\u4e2a\u6027\u8d28sn\uff0cs2n-sn\uff0cs3n-s2n

\u82e5an\u4e3a\u7b49\u5dee\u6570\u5217
\u5219Sn=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\u4ee5sn s2n-sn s3n-s2n \u4e5f\u662f\u7b49\u5dee\u6570\u5217 \u516c\u5dee\u4e3an*2d
\u8bbe\u7b49\u6bd4\u6570\u5217{an}\u7684\u516c\u6bd4\u4e3aq,
\u5219Sn,S2n-Sn,S3n-S2n\u6210\u7b49\u6bd4\u6570\u5217,\u516c\u6bd4\u4e3aq^n.
\u8bc1\u660e:\u5148\u8bc1\u660e\u4e00\u4e2a\u66f4\u4e00\u822c\u7684\u901a\u9879\u516c\u5f0f.\u5728\u7b49\u6bd4\u6570\u5217\u4e2d,
an=a1q^(n-1)
am=a1q^(m-1)
\u4e24\u5f0f\u76f8\u9664\u5f97an/am=q^(n-m),\u2234an=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
\u2234(S2n-Sn)/Sn=q^n.
\u540c\u7406,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.
\u2234(S3n-S2n)/(S2n-Sn)=q^n.
\u2234(S2n-Sn)/Sn=(S3n-S2n)/(S2n-Sn).\u5373(S2n-Sn)^2=Sn(S3n-S2n). \u6240\u4ee5sn s2n-sn s3n-s2n \u662f\u7b49\u6bd4\u6570\u5217 \u6253\u5b57\u8f9b\u82e6 \u671b\u91c7\u7eb3

\u8bc1\u660e\uff1a\u8bbe\u7b49\u5dee\u6570\u5217an\u7684\u9996\u9879\u4e3aa1\uff0c\u516c\u5dee\u4e3ad\uff0c\u5219Sn=a1+a2+\u2026+an\uff0cS2n-Sn=an+1+an+2+\u2026+a2n=a1+nd+a2+nd+\u2026+an+nd=Sn+n2d\uff0c\u540c\u7406\uff1aS3n-S2n=a2n+1+a2n+2+\u2026+a3n=an+1+an+2+\u2026+a2n+n2d=S2n-Sn+n2d\uff0c\u22342\uff08S2n-Sn\uff09=Sn+\uff08S3n-S2n\uff09\uff0c\u2234Sn\uff0cS2n-Sn\uff0cS3n-S2n\u662f\u7b49\u5dee\u6570\u5217\uff0e

证：
S(2n)-Sn=a(n+1)+a(n+2)+...+a(n+n)
=a1+nd+a2+nd+...+an+nd
=(a1+a2+...+an)+(nd+nd+...+nd)
=Sn+n²d
S(3n)-S(2n)=a(2n+1)+a(2n+2)+...+a(2n+n)
=a1+2nd+a2+2nd+...+an+2nd
=(a1+a2+...+an)+(2nd+2nd+...+2nd)
=Sn+2n²d
[S(2n)-Sn]-Sn=Sn+n²d-Sn=n²d
[S(3n)-S(2n)]-[S(2n)-Sn]=Sn+2n²d-(Sn+n²d)=n²d

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