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

an=a1*q^(n-1)
S4+2S2=3S3

S4-S3=2(S3-S2)

a4=2a3
a1q^3=2a1q^2
q=2
a4-a2=12
a1*q^3-a1*q=12
8a1-2a1=12
a1=2
an=2^n
nbn+1-(n+1)bn=n(n+1)
bn+1/(n+1)-bn/n=1
Xn=bn/n
b1=1
X1=1
Xn+1-Xn=1
Xn=n
bn/n=n
bn=n^2
Cn=1/[n(n+2)](n=2k-1)
C2k-1=[1/(2k-1)-1/(2k+1)]/2
奇数项的和：X=[1-1/(2n+1)]/2=n/(2n+1)
Cn=n2^(n-1)(n=2k)
C2k=2k*2^(2k-1)=k*2^2k
偶数项的和：Y=2*2^1+4*2^3+...+2n*2^(2n-1)
=1*2^2+2*2^4+3*2^6+...+n*2^2n
=4+2*4^2+3*4^3+...+n*4^n
=(4+4^2+4^3+...+4^n)+4^2+2*4^3+3*4^4+...+(n-1)*4^n
=(4+4^2+4^3+...+4^n)+4(4+2*4^2+3*4^3+...+(n-1)4^(n-1))
=(4+4^2+4^3+...+4^n)+4(Y-n*4^n)
=n(4^n-1)/3+4Y-4n*4^n
Y=n(1+11*4^n)/9
T2n=X+Y=n/(2n+1)+n(1+11*4^n)/9

Answer 3

(1)
an = a1.q^(n-1)
Sn = a1+a2+...+an
a4-a2=12
a1q(q^2- 1)=12 (1)
S4+2S2 = 3S3
a1(1-q^4) +2a1(1-q^2) = 3a1(1-q^3)
1-q^4 +2+2q^2 =3-3q^3
q^4-3q^3 +2q^2=0
q^2-3q+2=0
(q-1)(q-2)=0
q=2
from (1)
a1q(q^2- 1)=12
6a1=12
a1=2
an =2^n
nb(n+1) - (n+1)bn =n(n+1)
b(n+1)/(n+1) - bn/n =1
=> {bn/n} 是等差数列 , d=1
bn/n - b1/1 = n-1
bn/n - 1 = n-1
bn = n^2
(2)
log<2>an /[n^2.(n+2)] = 1/[n(n+2)]
2√bn/an = 2n/2^n = n. (1/2)^(n-1)
ie
cn
=1/[n(n+2)] ; n=2k-1
=n. (1/2)^(n-1) ; n=2k
n is odd
cn
= 1/[n(n+2)]
=(1/2) [ 1/n -1/(n+2) ]
c1+c3+...+c(2n-1)
=(1/2)[ ( 1- 1/3) + (1/3-1/5)+....+ (1/(2n-3) - 1/(2n-1)) ]
=(1/2) [ 1 - 1/(2n-1) ]
=n/(2n-1)
let
S = 2.(1/2)^1 + 4.(1/2)^3+.....+(2n).(1/2)^(2n-1) (3)
(1/4)S = 2.(1/2)^3 + 4.(1/2)^5+.....+(2n).(1/2)^(2n+1) (4)
(3)-(4)
(3/4)S = 2[ 1/2^1 + 1/2^3+...+1/2^(2n-1) ] - (2n).(1/2)^(2n+1)
= (2/3) [ 1 - (1/2)^(2n) ] - (2n).(1/2)^(2n+1)
= (2/3) -(n+ 2/3).(1/2)^(2n)
S =8/9 - (3/4)(n+ 2/3).(1/2)^(2n)
n is even
cn
=n. (1/2)^(n-1)
c2+c4 + c6+....+ c(2n)
=S
=8/9 - (3/4)(n+ 2/3).(1/2)^(2n)
T(2n)
= [c1+c3+...+c(2n-1) ] +[c2+c4+...+c(2n)]
=n/(2n-1) +8/9 - (3/4)(n+ 2/3).(1/2)^(2n)