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sn s2n%sn s3n%s2n成等差

证明:设等差数列an的首项为a1,公差为d,则Sn=a1+a2+…+an,S2n-Sn=an+1+an+2+…+a2n=a1+nd+a2+nd+…+an+nd=Sn+n2d,同理:S3n-S2n=a2n+1+a2n+2+…+a3n=an+1+an+2+…+a2n+n2d=S2n-Sn+n2d,∴2(S2n-Sn)=Sn+(S3n-S

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=nd,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=nd,所以从Sn开始就是等差.

这个很简单啊,只要知道sn=na1+n(n-1)/2*(d),带入求出sn s2n,s3n就很快证明除了

sn=a1n+n(n-1)d/2 s2n=2a1n+n(2n-1)d s3n=3a1n+3n(3n-1)d/2 s2n-sn-sn=2a1n+n(2n-1)d-2a1n+n(n-1)d=n^2d s3n-s2n-(s2n-sn)=s3n-2s2n+sn=3a1n+3n(3n-1)d/2-4a1n-2n(2n-1)d+a1n+n(n-1)d/2=n^2d 所以,sn,s2n-sn,s3n-s2n成等差数列

不能.因为把an到a2n之间的数打乱,这不影响S2n-Sn的结果,但是an的顺序乱了,所以不是等差.

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; 公差:(S2n-Sn)-Sn=(

等差数列

他们就是等差数列啊

反证法: Sn=a1+a2+……+an; S2n-Sn=an+1+an+2+……+a2n; S3n-S2n=a2n+1+a2n+2+……+a3n; 如果{an}不是等差数列, an+1-a1≠nd且a2n+1-an+1≠nd;以此类推可得S2n-Sn-Sn≠n^2*d;S3n-S2n-(S2n-Sn)≠n^2*d,这与已知相矛盾,所以{an}是等差数列

等差数列中,Sn、S2n-Sn、S3n-S2n的公差为n^2*d 等比数列中,Sn、S2n-Sn、S3n-S2n的公比为q^n

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