Question

数列题，可以只写第一问（高手帮忙写下第二问）

Answer 1

追答

第一题

追问

能写第二题不

追答

用了错位相减法

Answer 2

(I)
a1=1, a2=2
a(n+2)=( 1+ [cos(nπ/2)]^2).an + [sin(nπ/2)]^2
if n is odd
a(n+2) = an +1
a(n+2) -an =1
an - a1 = (n-1)/2
an = (n+1)/2

if n is even
a(n+2) = 2an
an = 2^[(n-2)/2] .a2
= 2^(n/2)
ie
an = (n+1)/2 , if n is odd
= 2^(n/2) , if n is even
(II)
let
S= 1.(1/2)^1+2.(1/2)^2+...+n.(1/2)^n (1)
(1/2)S= 1.(1/2)^2+2.(1/2)^3+...+n.(1/2)^(n+1) (2)
(1)-(2)
(1/2)S = (1/2+1/2^2+...+1/2^n) -n.(1/2)^(n+1)
= 1- (1/2)^n -n.(1/2)^(n+1)
S = 2- [n+2](1/2)^n

bn = a(2n-1)/a(2n)
= n/2^n
Sn =b1+b2+...+bn
= S
=2- [n+2](1/2)^n
<2

追问

写在纸上行不

追答

不行