等比数列{an}的前n项和为Sn,已知对任意的n∈N+,点(n,Sn)均在函数y+b^x+r(b>0)且b≠1,b,r均为常数)的图像上.(1)求r的值;(2)当b=2时,记bn=n/2an(n∈N+)求数列{bn}的前n项的Tn(3)当b=3时,记Cn=2an/
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![等比数列{an}的前n项和为Sn,已知对任意的n∈N+,点(n,Sn)均在函数y+b^x+r(b>0)且b≠1,b,r均为常数)的图像上.(1)求r的值;(2)当b=2时,记bn=n/2an(n∈N+)求数列{bn}的前n项的Tn(3)当b=3时,记Cn=2an/](/uploads/image/z/3929245-61-5.jpg?t=%E7%AD%89%E6%AF%94%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%A5%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84n%E2%88%88N%2B%2C%E7%82%B9%EF%BC%88n%2CSn%EF%BC%89%E5%9D%87%E5%9C%A8%E5%87%BD%E6%95%B0y%2Bb%5Ex%2Br%28b%3E0%29%E4%B8%94b%E2%89%A01%2Cb%2Cr%E5%9D%87%E4%B8%BA%E5%B8%B8%E6%95%B0%EF%BC%89%E7%9A%84%E5%9B%BE%E5%83%8F%E4%B8%8A.%EF%BC%881%EF%BC%89%E6%B1%82r%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89%E5%BD%93b%3D2%E6%97%B6%2C%E8%AE%B0bn%3Dn%2F2an%28n%E2%88%88N%2B%29%E6%B1%82%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E7%9A%84Tn%EF%BC%883%EF%BC%89%E5%BD%93b%3D3%E6%97%B6%2C%E8%AE%B0Cn%3D2an%2F)
等比数列{an}的前n项和为Sn,已知对任意的n∈N+,点(n,Sn)均在函数y+b^x+r(b>0)且b≠1,b,r均为常数)的图像上.(1)求r的值;(2)当b=2时,记bn=n/2an(n∈N+)求数列{bn}的前n项的Tn(3)当b=3时,记Cn=2an/
等比数列{an}的前n项和为Sn,已知对任意的n∈N+,点(n,Sn)均在函数y+b^x+r(b>0)且b≠1,b,r均为常数)的图像上.
(1)求r的值;
(2)当b=2时,记bn=n/2an(n∈N+)求数列{bn}的前n项的Tn
(3)当b=3时,记Cn=2an/(an+1)(3an+1),求证:C1+C2+...+Cn
等比数列{an}的前n项和为Sn,已知对任意的n∈N+,点(n,Sn)均在函数y+b^x+r(b>0)且b≠1,b,r均为常数)的图像上.(1)求r的值;(2)当b=2时,记bn=n/2an(n∈N+)求数列{bn}的前n项的Tn(3)当b=3时,记Cn=2an/
(1)
点(n,Sn)均在函数y=b^x+r
n=1,a1=b+r (1)
n=2,
S2= b^2 +r
a2+(b+r)=b^2 +r
a2 = b(b-1) (2)
n=3,
S3 =b^3+r
a3+ b^2 +r =b^3+r
a3= b^2(b-1) (3)
a3/a2 = a2/a1
b^2(b-1)/[b(b-1)] = b(b-1)/(b+r)
b(b+r) =b(b-1)
br= -b
r= -1
(2)
b=2
Sn=2^n-1
an = Sn-S(n-1) = 2^(n-1)
bn=(n/2)an
= (1/2)(n.2^(n-1) )
consider
1+x+x^2+..+x^n = (x^(n+1)- 1)/(x-1)
1+2x+..+nx^(n-1) =[(x^(n+1)- 1)/(x-1)]'
= [nx^(n+1) - (n+1)x^n + 1]/(x-1)^2
put n=2
summation(1:1->n)i.2^(i-1)
=n.2^(n+1) - (n+1).2^n + 1
= 1+ (n-1).2^n
bn=(n/2)an
= (1/2)(n.2^(n-1) )
Tn=b1+b2+...+bn
=(1/2){summation(1:1->n)i.2^(i-1)}
=(1/2)[1+ (n-1).2^n]
(3)
b=3
Sn=3^n-1
an= Sn-S(n-1) = 2.3^(n-1)
cn = 2an/(an+1)(3an+1)
= 4.3^(n-1) /[( 1+2.3^(n-1)).(1+ 2.3^n) ]
= 1/( 1+2.3^(n-1)) - 1/(1+ 2.3^n)
c1+c2+...+cn
= 1/3 - 1/(1+ 2.3^n)