s2n-sn成等差数列
@戈叙1224：等差数列性质 为什么S2n - Sn=an+1 +an+2 +....+a2n S3n - S2n=a2n+1 +a2n+2 +...+a3n 公差为 -
生非18177267120…… S(2n)=a1+a2+...+a(2n) (1) Sn = a1+a2+....+an (2) S(3n) = a1+a2+...+a(3n) (3) (1)-(2) S(2n)-Sn=an+1 +an+2 +....+a2n (3)-(1) S(3n)-S(2n)=a(2n+1) +a(2n+2) +...+a(3n)
@戈叙1224：等差数列前N项和性质 -
生非18177267120…… Sn,S2n-Sn,S3n-S2n..........成等差数列,公差为n^2*d 证明如下: Sk=ka1+k(k-1)d/2 S2k=2ka1+2k(2k-1)d/2 S3k=3ka1+3k(3k-1)d/2 S2k-Sk=ka1+k(3k-1)d/2 S3k-S2k=ka1+k(5k-1)d/2 (S2k-Sk)-Sk=k^2*d (S3k-S2k)-(S2k-Sk)=k^2*d 所以 等差数列依...
@戈叙1224：证明:Sn,S2n - Sn,S3n - S2n,…成等差数列. - 作业帮
生非18177267120…… [答案] 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...
@戈叙1224：设Sn为等差数列{an}的前n项和,证明Sn,S2n - Sn,S3n - S2n成等差数列. - 作业帮
生非18177267120…… [答案] 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...
@戈叙1224：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___. - 作业帮
生非18177267120…… [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...
@戈叙1224：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? - 作业帮
生非18177267120…… [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=...
@戈叙1224：请问等差数列的前N项和有什么性质 等比数列呢? -
生非18177267120…… 等差S(2n-1)=(2n-1)an sn s2n-sn s3n-s2n 成等差数列 sn-s(n-1)=an
@戈叙1224：等差数列{an}的前n项和为sn,若sn=30,s2n=100,则s3n=( ) -
生非18177267120…… 等差数列的sn,s2n-sn,s3n-s2n也成等差数列 2(s2n-sn)=sn+s3n-s2n 2(100-30)=30+s3n-100 2*70=s3n-70 140=s3n-70 s3n=210
@戈叙1224：若a与b的夹角为,且为钝角,则a·b在等差数列中,sn,s2n - sn,s3n=s2n成等差,证明一下为什么? - 作业帮
生非18177267120…… [答案] 若a与b的夹角为θ,且为钝角.则cosθ0,|b|>0 所以: a·b=|a||b|cosθ
@戈叙1224：已知等差数列{an}的前n项和为Sn,公差为d.求证:Sn,S2n - Sn,S3n - S2n,.成等差数列 - 作业帮
生非18177267120…… [答案] Sn=a1n+n(n-1)d/2S2n=2a1n+n(2n-1)dS3n=3a1n+3n(3n-1)d/2S2n-Sn-Sn=2a1n+n(2n-1)d-2a1n+n(n-1)d=n^2dS3n-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...
生非18177267120…… S(2n)=a1+a2+...+a(2n) (1) Sn = a1+a2+....+an (2) S(3n) = a1+a2+...+a(3n) (3) (1)-(2) S(2n)-Sn=an+1 +an+2 +....+a2n (3)-(1) S(3n)-S(2n)=a(2n+1) +a(2n+2) +...+a(3n)
@戈叙1224：等差数列前N项和性质 -
生非18177267120…… Sn,S2n-Sn,S3n-S2n..........成等差数列,公差为n^2*d 证明如下: Sk=ka1+k(k-1)d/2 S2k=2ka1+2k(2k-1)d/2 S3k=3ka1+3k(3k-1)d/2 S2k-Sk=ka1+k(3k-1)d/2 S3k-S2k=ka1+k(5k-1)d/2 (S2k-Sk)-Sk=k^2*d (S3k-S2k)-(S2k-Sk)=k^2*d 所以 等差数列依...
@戈叙1224：证明:Sn,S2n - Sn,S3n - S2n,…成等差数列. - 作业帮
生非18177267120…… [答案] 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...
@戈叙1224：设Sn为等差数列{an}的前n项和,证明Sn,S2n - Sn,S3n - S2n成等差数列. - 作业帮
生非18177267120…… [答案] 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...
@戈叙1224：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___. - 作业帮
生非18177267120…… [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...
@戈叙1224：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? - 作业帮
生非18177267120…… [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=...
@戈叙1224：请问等差数列的前N项和有什么性质 等比数列呢? -
生非18177267120…… 等差S(2n-1)=(2n-1)an sn s2n-sn s3n-s2n 成等差数列 sn-s(n-1)=an
@戈叙1224：等差数列{an}的前n项和为sn,若sn=30,s2n=100,则s3n=( ) -
生非18177267120…… 等差数列的sn,s2n-sn,s3n-s2n也成等差数列 2(s2n-sn)=sn+s3n-s2n 2(100-30)=30+s3n-100 2*70=s3n-70 140=s3n-70 s3n=210
@戈叙1224：若a与b的夹角为,且为钝角,则a·b在等差数列中,sn,s2n - sn,s3n=s2n成等差,证明一下为什么? - 作业帮
生非18177267120…… [答案] 若a与b的夹角为θ,且为钝角.则cosθ0,|b|>0 所以: a·b=|a||b|cosθ
@戈叙1224：已知等差数列{an}的前n项和为Sn,公差为d.求证:Sn,S2n - Sn,S3n - S2n,.成等差数列 - 作业帮
生非18177267120…… [答案] Sn=a1n+n(n-1)d/2S2n=2a1n+n(2n-1)dS3n=3a1n+3n(3n-1)d/2S2n-Sn-Sn=2a1n+n(2n-1)d-2a1n+n(n-1)d=n^2dS3n-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...
