s2n-1+2n-1+an推导

籍艺威2346在等差数列中,an不等于0,a(n - 1) - an^2+a(n+1)=0(n>=2),若S2n - 1=38,则n=? -
宿房须18311236770 ______ an等差,所以a(n-1)+a(n+1)=2an 所以2an=an^2 an不等于0,所以an=2 S2n-1=2*(2n-1)=382n-1=19 n=10

籍艺威2346如何推导此公式? -
宿房须18311236770 ______ 虽然你没说,我猜应该是等差数列吧.. 设An=a+(n-1)*b,Bn=c+(n-1)*d. 则:S2n-1=[2a+(2n-2)b]*n/2=[a+(n-1)b]*n=An*n T2n-1=[2c+(2n-2)d]*n/2=[c+(n-1)d]*n=Bn*n 所以:S2n-1/T2n-1=An/Bn 拿分来,嘿嘿..

籍艺威2346等比数列的性质与等差数列的性质 -
宿房须18311236770 ______ 等差数列 通项公式 an=a1+(n-1)d an=Sn-S(n-1) (n≥2) an=kn+b(k,b为常数) 前n项和 倒序相加法推导前n项和公式: Sn=a1+a2+a3······+an =a1+(a1+d)+(a1+2d)+······+[a1+(n-1)d] ① Sn=an+(an-d)+(an-2d)+······+...

籍艺威2346等差数列的项数为2N - 1 -
宿房须18311236770 ______ S奇=a1+a3+a5……+a2N-3+a2N-1 S偶=a2+a4+a6……+a2N-2 S奇-S偶=(a1-a2)+(a3-a4)+……(a2N-3-a2N-2)+a2N-1 =(-d)+(-d)+(-d)……+(-d)+a2N-1 =a2N-1-(N-1)d=aN S奇+S偶=S2N-1=(a1+a2N-1)*(2N-1)/2=(2N-1)aN 所以S奇=NaN S偶=(N-1)aN 所以SS奇/S偶=N/N-1

籍艺威2346S2n - 1=an(2n - 1)推理过程 -
宿房须18311236770 ______[答案] S2n-1=(a1+a2n-1)*(2n-1)/2 =2an/2*(2n-1)=an(2n-1)

籍艺威2346等差数列S2n - S2n - 2+a3+a2为什么等于a2n+a2n - 1+a3+a2? -
宿房须18311236770 ______ S2n-S2n-2 S2n是前2n项的和.S2n-2是前2n-2项的和.他们之差就是第2n项和第2n-1项的和 也就是a2n+a(2n-1) 后面的a3和a2照写 如仍有疑惑,欢迎追问. 祝:学习进步!

籍艺威2346在各项均不为零的等差数列{an}中 ,若an+1 - an∧2+an - 1=0则s2n - 1 - 4n= -
宿房须18311236770 ______ 因为an=2 根据等差数列奇数项和公式S2n-1=(2n-1)an可得 S2n-1-4n=2*(2n-1)-4n=-2

籍艺威2346等差数列的前N项和Sn,若S2n - 1=(2n - 1)(2n+1),则Sn=() -
宿房须18311236770 ______[答案] S2n-1=(2n-1)(2n+1) S2n-1=(2n-1)(2n-1+2) Sn=n(n+2)=n2+2n

籍艺威2346等差数列an与S2n - 1的关系怎么会是an=(S2n - 1)/(2n - 1)? -
宿房须18311236770 ______[答案] 2*S2n-1 =(a1+a2n-1)+(a2+a2n-2)+……+(a2n-1+a1) =(2n-1)*2an => an=(S2n-1)/(2n-1)