Question

高二数列题

Answer 1

anan+1=（1/3）? a1a2=1/3得a2=1/6 lgan+1=nlg1/3-lgan=nlg1/3-（（n-1）lg1/3-lgan-1）=lg1/3+lgan-1 故奇数项lga2m+1=lg1/3+lga2m-1=mlg1/3+lga1 a2m+1=2（1/3）∧m 偶数项lga2m=lg1/3+lga2m-2=（m-1）lg1/3+lga2 a2m=1/2（1/3）∧m 奇数项和=2/（1-1/3）=3，偶数项和=1/6/（1-1/3）=1/4 无穷数列cn和=2（奇数项和+偶数项和）-a1=2（3+1/4）-2=9/2

Answer 2

图

Answer 3

(1)
an=a1+(n-1)d
Sn = a1+a2+...+an
S10=110
5(2a1+9d)=110
2a1+9d=22 (1)
S15=240
15(a1+7d) =240
a1+7d=16 (2)
2(2)-(1)
5d= 10
d=2
from (2)
a1+14=16
a1=2
an = 2+2(n-1) = 2n
(2)
bn
= a(n+1) /an + an/a(n+1)
=(n+1)/n + n/(n+1)
= 1+ 1/n + 1 - 1/(n+1)
=2 + [1/n - 1/(n+1) ]
Tn
=b1+b2+...+bn
=2n + [ 1 - 1/(n+1) ]
=2n + [ n/(n+1)]
(3)
let
S = 1.(1/2)^0 +2.(1/2)^1+....+n.(1/2)^(n-1) (1)
(1/2)S = 1.(1/2)^1 +2.(1/2)^2+....+n.(1/2)^n (2)
(1)-(2)
(1/2)S = [1+1/2+...+(1/2)^(n-1) ] - n.(1/2)^n
= 2( 1- 1/2^n) -n.(1/2)^n
S = 4( 1- 1/2^n) -2n.(1/2)^n
cn
= an/2^n
= 2n/2^n
= n .(1/2)^(n-1)
Pn
=c1+c2+...+cn
=S
=4( 1- 1/2^n) -2n.(1/2)^n
=4 -(2n+4).(1/2)^n

Answer 4