Question

请问数学大神17题怎么做?

Answer 1

(17)
(1)
a1=0
Sn=a1+a2+...+an
Sn = 2a(n+1) +n
for n>=2
an = Sn -S(n-1)
= 2a(n+1) -2an +1
a(n+1) = (3/2)an -1/2
a(n+1) - 1 = (3/2)(an - 1)
=> {an - 1} 是等比数列, q=3/2
an - 1 =(3/2)^(n-1) .(a1 - 1)
an = 1-(3/2)^(n-1)
(2)
bn=2+2(n-1)=2n
let
S = 1.(3/2)^1 +2.(3/2)^2+...+n.(3/2)^n (1)
(3/2)S = 1.(3/2)^2 +2.(3/2)^3+...+n.(3/2)^(n+1) (2)
(2)-(1)
(1/2)S =n.(3/2)^(n+1) -[ (3/2)^1 +(3/2)^2 +...+(3/2)^n ]
=n.(3/2)^(n+1) - 3[ (3/2)^n -1 ]
S =2n.(3/2)^(n+1) - 3[ (3/2)^n -1 ]
cn
= a(n+1).bn
=[ 1-(3/2)^n ] . 2n
=2n - 2[n.(3/2)^n]
Tn
=c1+c2+...+cn
=n(n+1) - 2S
=n(n+1) - 2{ 2n.(3/2)^(n+1) - 3[ (3/2)^n -1 ] }
=n(n+1) - 4n.(3/2)^(n+1) + 6[ (3/2)^n -1 ]
=n^2+n -6 - (6n-6). (3/2)^n