Question

请问这道数列题怎么做？

Answer 1

如图所示

Answer 2

如图所示

Answer 3

(1)
an.a(n+1)= (2^n)^2
n=1
a1.a2 = 2^2
a2 =4
n=2
a2.a3 = 2^4
a3 = 4
n=3
a3.a4 =2^6
a4= 16
(2)
an.a(n+1)= (2^n)^2
an.a(n+1) = 2^(2n)
lnan + ln[a(n+1)] =2nln2
let
ln[a(n+1)] + k1.(n+1) + k2 = -(lnan + k1.n + k2)
coef. of n
-2k1 = 2ln2
k1 = -ln2
coef. of constant
-2k2 - k1 = 0
-2k1 +ln2 =0
k2 =(1/2)ln2
ie
ln[a(n+1)] - (ln2)(n+1) + (1/2)ln2 = -[lnan + (ln2)n + (1/2)ln2]
=> {lnan + (ln2)n + (1/2)ln2} 是等比数列，q=-1
lnan + (ln2)n + (1/2)ln2 =(-1)^(n-1) .[lna1 + ln2 + (1/2)ln2 ]
=(3/2)ln2 .(-1)^(n-1)
lnan = -(ln2)n + (1/2)ln2 + (3/2)ln2 .(-1)^(n-1)
an = e^[ -(ln2)n + (1/2)ln2 + (3/2)ln2 .(-1)^(n-1) ]
= (1/2)^n . 2^(1/2) . [2^(3/2)]^[(-1)^(n-1)]
a2n= (1/2)^(2n) . 2^(1/2) . 2^(-3/2)
= (1/2)(1/2)^(2n)
Sn
=a2+a4+....+a(2n)
=(1/8) [ 1 - 1/2^(2n) ]/( 1 - 1/4)
=(1/6) [ 1 - 1/2^(2n) ]

Answer 4