sn+s2n-sn的公差
答:(Sn) (S2n-Sn) (S3n-S2n)成公差为n²d的等差数列
答:方法一:对于这类题,我认为首先要明确的是:等差数列中Sn、S2n-Sn、S3n-S2n……也成等差数列 以下是证明过程:设等差数列{an}的首项为a1,公差为d Sn=a1+a2+……+an S2n-Sn=a(n+1)+a(n+2)+……+a(2n)=a1+nd+a2+nd+……+an+nd =a1+a2+……+an+n^2d =Sn+n^2d 同理:S3n-S2n...
答:等差数列的每n项组合,组成一个新的数列,同样也是等差数列,只是新数列的增量是原数列增量的n倍,设新数列的增量为x,则200+3x=500,解得x=100,则后面3n项的和为300+(3+4+5)x=1500.故答案为:1500.或者 由题意有Sn=100,S3n-Sn=500.又Sn,S2n-Sn,S3n-S2n成等差数列,其公差为...
答:Sn,S2n-Sn,S3n-S2n显然是等差数列(公差为nd)∴2(S2n-Sn﹚=Sn+S3n-S2n 2﹙100-30)=30+s3n-100 ∴S3n=210 这个等差数列前3n项的和为210
答:S2n为该数列的前2n项和,S3n为该数列的前3n项和,则Sn,S2n-Sn,S3n-S2n也为等差数列。2、如果一个数列从第二项起,每一项与它的前一项的差等于同一个常数,这个数列就叫做等差数列,而这个常数叫作等差数列的公差,公差常用字母d表示。例如:1,3,5,7,9…...(2n-1)。
答:巧算:2(S2n-Sn)=Sn+S3n-S2n 2*75=25+S3n-100 S3n=225 定义算法:Sn=a1+a2+...+an=25 S2n=a1+a2+...+an+a(n+1)+a(n+2)+...+a(n+n)=a1+a2+...+an+a1+nd+a2+nd+...+an+nd =2(a1+a2+...+an)+n²d=2×25+n²d=100 n²d=50 a(n+1)+...
答:相关性质 1、常数列:C,C,…,C是公差d=0的等差数列。2、等差中项:如果a,A,b成等差数列,则A叫作a与b的等差中项,且A=(a+b)/2。3、若Sn是等差数列的前n项和,则Sn,S2n-Sn,nS3n-S2n,…是一个等差数列。4、若{an}是等差数列,公差d>0时{an}是递增数列,d<0时{an}是...
答:Sn,S2n-Sn,……成等差数列,等差数列前n项和知道公式吧,公差就等于S2n-Sn-Sn,结果就是n²d
答:Sn,S2n-Sn,S3n-S2n,……仍成等差数列 公差为nd S3,S6-S3,S9-S6,S12-S9成等差数列,公差为3d 所以S6-S3-S3=3d S6=2S3+3d S6,S12-S6也成等差,公差为6d 所以S12-S6-S6=6d S12=2S6+6d S3:S6=1/3 S3/(2S3+3d)=1/3 S3=3d 所以S6=9d S12=24d S6:S12=3/8 ...
答:则Sn,S2n -Sn ,S3n -S2n,…仍然成等差数列,公差为k^2d .(4)若数列{an}与{bn}均为等差数列,且前n项和分别是Sn和Tn,则am/bm=S2m-1/T2m-1。⑸在等差数列中,S = a,S = b (n>m),则S = (a-b).⑹等差数列中, 是n的一次函数,且点(n, )均在直线y = x + (...
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逯桂13131651225:
等差、等比数列中,Sn、S2n - Sn、S3n - S2n...的公差和公比都怎么表示? -
33778周欢
: 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
逯桂13131651225:
等差数列的前n项和也构成一个等差数列,即Sn,S2n - Sn,S3n - S2n,…为等差数列,公差为___. -
33778周欢
:[答案] 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,∴(S2n-Sn)-Sn)=n2d,同理:S3n-S2n=a2n+1+a2n+2+…+a3n=an+1+an+2+…+a2n+n2d=S2n-Sn+n2d,∴(S3n...
逯桂13131651225:
等差数列的性质{a(n)}为等差数列,公差为d,则Sn,S2n - Sn,S3n - S2n,也为等差数列,其公差为多少? -
33778周欢
:[答案] a(n)=a+(n-1)d,S(n)=na+n(n-1)d/2,S(2n)=2na+2n(2n-1)d/2,S(3n)=3na+3n(3n-1)d/2.S(3n)-S(2n)=na+nd[9n-3-4n+2]/2=na+nd[5n-1]/2,S(2n)-S(n)=na+dn[4n-2-n+1]/2=na+nd[3n-1]/2,[S(3n)-S(2n)]-[S(2n)-S(n)]=nd[2n]/2=...
逯桂13131651225:
等差数列an的前n项和为Sn,若Sn=25,S2n=100,则S3n=? -
33778周欢
: Sn,S2n-Sn,S3n-S2n是等差数列2(S2n-Sn)=Sn+S3n-S2n2*75=25+S3n-100 S3n=225
逯桂13131651225:
两组数列分别为等差数列和等比数列它们的和为SN则两组数列SN,S2N - SN,S3N - S2N,成不成等差等比,如果成的话为什么成,公差公比又是什么 -
33778周欢
:[答案] 两组数列分别为等差数列和等比数列它们的和为SN 则两组数列SN,S2N-SN,S3N-S2N,成等差或等比 (1)若原数列是等差数列,则新数列是等差数列,公差是N²*d (2)若原数列是等比数列,则新数列是等比数列,公比是q^N
逯桂13131651225:
若an是等差数列,则Sn,S2n - Sn,S3n - S2n,…也成等差数列,求公差公差为dn^2 求证明过程 详细点 -
33778周欢
:[答案] 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] -...
逯桂13131651225:
等差数列前n项和为sn,则sn,s2n - sn,s3n - s2n也成等差. -
33778周欢
: 证明可知sn+(s3n-s2n)=2(s2n-sn) 所以他们组成等差数列
逯桂13131651225:
等差数列的性质 -
33778周欢
: a(n)=a+(n-1)d,S(n)=na+n(n-1)d/2,S(2n)=2na+2n(2n-1)d/2,S(3n)=3na+3n(3n-1)d/2.S(3n)-S(2n)=na+nd[9n-3-4n+2]/2=na+nd[5n-1]/2,S(2n)-S(n)=na+dn[4n-2-n+1]/2=na+nd[3n-1]/2,[S(3n)-S(2n)]-[S(2n)-S(n)]=nd[2n]/2=dn^2,[S(2n)-S(n)]-[S(n)]=nd[3n-1-n+1]/2=dn^2.所以,Sn,S2n-Sn,S3n-S2n,为等差数列,其公差为dn^2.
逯桂13131651225:
数列2(S2n - Sn)为什么等于Sn+S3n - S2n 怎么来的,求详解 -
33778周欢
:[答案] 设公差为d 2[S(2n)-Sn]=2[a1+a2+...+a(2n)-a1+a2+...+an] =2[a(n+1)+a(n+2)+...+a(2n)] =2[a1+nd+a2+nd+...+an+nd] =2[(a1+a2+...+an)+n²d] Sn+S(3n)-S(2n) =a1+a2+...+an+a1+a2+...+a(3n)-[a1+a2+...+a(2n)] =a1+a2+...+an+a(2n+1)+a(2n+2)+...+a(3n) ...
逯桂13131651225:
已知等差数列{an}的前n项和为Sn,公差为d. -
33778周欢
: {an}为等差数列,首项为a1,则 Sn=na1+n(n-1)d/2 S2n=2na1+2n(2n-1)d/2 因为,S2n-Sn=na1+n(3n-1)d/2 所以,(S2n-Sn)-Sn=n²d 因为,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...