设数列{an}的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n(c为常数,c不等于1,n属于正整数)设数列{an}的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n,且a1,a2,a3成等差数列.c=2,an=n若数列{bn}是首项为1,公比为c的等比数列,记A
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![设数列{an}的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n(c为常数,c不等于1,n属于正整数)设数列{an}的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n,且a1,a2,a3成等差数列.c=2,an=n若数列{bn}是首项为1,公比为c的等比数列,记A](/uploads/image/z/5126207-23-7.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A5S1%3D1%2CSn%2B1%2FSn%3Dn%2Bc%2Fn%EF%BC%88c%E4%B8%BA%E5%B8%B8%E6%95%B0%2Cc%E4%B8%8D%E7%AD%89%E4%BA%8E1%2Cn%E5%B1%9E%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A5S1%3D1%2CSn%2B1%2FSn%3Dn%2Bc%2Fn%2C%E4%B8%94a1%2Ca2%2Ca3%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97.c%3D2%2Can%3Dn%E8%8B%A5%E6%95%B0%E5%88%97%7Bbn%7D%E6%98%AF%E9%A6%96%E9%A1%B9%E4%B8%BA1%2C%E5%85%AC%E6%AF%94%E4%B8%BAc%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E8%AE%B0A)
设数列{an}的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n(c为常数,c不等于1,n属于正整数)设数列{an}的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n,且a1,a2,a3成等差数列.c=2,an=n若数列{bn}是首项为1,公比为c的等比数列,记A
设数列{an}的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n(c为常数,c不等于1,n属于正整数)
设数列{an}的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n,且a1,a2,a3成等差数列.c=2,an=n
若数列{bn}是首项为1,公比为c的等比数列,记An=a1b1=a1b2+...+anbn,Bn=a1b1-a2b2+...+(-1)^n-1anbn,n属于正整数.证明An+3B2n=4/3(1-4^n)
设数列{an}的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n(c为常数,c不等于1,n属于正整数)设数列{an}的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n,且a1,a2,a3成等差数列.c=2,an=n若数列{bn}是首项为1,公比为c的等比数列,记A
a(1)=s(1)=1
s(n+1)/s(n)=(n+c)/n
s(2)/s(1)=1+c=s(2)=a(1)+a(2)=s(1)+a(2)=1+a(2),a(2)=c
2a(2)=a(1)+a(3),a(3)=2a(2)-a(1)=2c-1.
s(3)/s(2)=(2+c)/2=[1+c+2c-1]/(1+c)=(2+c)/2,6c=(2+c)(1+c)=2+3c+c^2
0=c^2-3c+2=(c-1)(c-2),c不等于1.
c=2
s(n+1)/s(n)=(n+2)/n
s(n+1)/(n+2)=s(n)/n
s(n+1)/[(n+2)(n+1)]=s(n)/[(n+1)n]=...=s(1)/[2*1]=1/2
s(n)=n(n+1)/2
s(n+1)=(n+1)(n+2)/2
a(n+1)=s(n+1)-s(n)=(n+1)[n+2-n]/2=n+1
a(n)=n
b(n)=2^(n-1)
A(n)=a(1)b(1)+a(2)b(2)+...+a(n)b(n)=1*1+2*2+3*2^2+...+(n-1)*2^(n-2)+n*2^(n-1)
2A(n)=1*2+2*2^2+3*2^3+...+(n-1)*2^(n-1)+n*2^(n)
A(n)=2A(n)-A(n)=-1-2-2^2-...-2^(n-1)+n*2^(n)
=n*2^(n)-[2^(n)-1]
=(n-1)*2^n+1
B(n)=a(1)b(1)-a(2)b(2)+...+(-1)^(n-1)a(n)b(n)=1*1-2*2+3*2^2+...+(n-1)*2^(n-2)*(-1)^(n-2)+n*2^(n-1)*(-1)^(n-1)
=1*1+2*(-2)+3*(-2)^2+...+(n-1)*(-2)^(n-2)+n*(-2)^(n-1)
-2B(n)=1*(-2)+2*(-2)^2+3*(-2)^3+...+(n-1)*(-2)^(n-1)+n*(-2)^n
3B(n)=B(n)-[-2B(n)]=1+(-2)+(-2)^2+...+(-2)^(n-1)-n*(-2)^n
=[1-(-2)^n]/(1+2)-n*(-2)^n
=[1-(-2)^n]/3 - n*(-2)^n
B(n)=[1-(-2)^n]/9-n*(-2)^n/3
yy