Question

高中数学数列题求解

Answer 1

（1）S4-S3=2*(S3-S2)
a4=2*a3
所以等比数列{an}的公比为2
a4-a2=4*a2-a2=3*a2=12，所以a2=4
所以an=a2*2^(n-2)=2^n
n*b(n+1)-(n+1)*bn=n*(n+1)，两边除以n(n+1)
b(n+1)/(n+1)-bn/n=1
又因为b1/1=b1=1
所以{bn/n}是以1为首项，1为公差的等差数列
bn/n=1+(n-1)*1=n
bn=n^2
（2）当n=2k-1时，cn=log(2,2^n)/[n^2*(n+2)]=1/n(n+2)
当n=2k时，cn=2√(n^2)/(2^n)=2n/(2^n)
T(2n)=[c1+c3+...+c(2n-1)]+[c2+c4+...+c(2n)]
={1/(1*3)+1/(3*5)+...+1/[(2n-1)*(2n+1)]}+[4/4+8/16+...+4n/(4^n)]
=[1-1/3+1/3-1/5+...+1/(n-2)-1/(2n+1)]+{1/(4^0)+2/(4^1)+...+n/[4^(n-1)]}
=[1-1/(2n+1)]+(1/3)*{4*{1/(4^0)+2/(4^1)+...+n/[4^(n-1)]}-{1/(4^0)+2/(4^1)+...+n/[4^(n-1)]}}
=2n/(2n+1)+(1/3)*{{4+2/(4^0)+...+n/[4^(n-2)]}-{1/(4^0)+2/(4^1)+...+n/[4^(n-1)]}}
=2n/(2n+1)+(1/3)*{4-n/[4^(n-1)]+{1/(4^0)+1/(4^1)+...+1/[4^(n-1)]}}
=2n/(2n+1)+(1/3)*{4-n/[4^(n-1)]+[1-1/(4^n)]/(1-1/4)]}
=2n/(2n+1)+(1/3)*{4-n/[4^(n-1)]+4/3-(1/3)/[4^(n-1)]}
=2n/(2n+1)+(1/3)*{16/3-(n+1/3)/[4^(n-1)]}
=2n/(2n+1)+16/9-(n/3+1/9)/[4^(n-1)]

Answer 2

(18)
(1)
a1=2
a(n+1)=3an-2
a(n+1) - 1 = 3( an -1)
=> { an -1} 是等比数列, q=3
an -1 = 3^(n-1) .(a1 -1)
=3^(n-1)
an =1+ 3^(n-1)
(2)
1/an = 1/[1+ 3^(n-1)] < 1/3^(n-1)
1/a1 = 1/(1+1) = 1/2
1/a1+1/a2+...+1/an
=1/2 +(1/a2+...+1/an)
< 1/2 + [1/3^1 + 1/3^2+...+1/3^(n-1) ]
=1/2 + (1/3)[ 1- 1/3^(n-1) ]/(1-1/3)
=1/2 + (1/2)[ 1- 1/3^(n-1) ]
< 1/2 +1/2
=1

追问

谢谢啦，但是我已经采纳别人了