Question

已知数列an前n项和为Sn，且满足4（n+1)(Sn+1)=(n+2)^2an(n属于正整数） 求an

只要做第2问求an就行

Answer 1

(2)
a1=8

4(n+1)(Sn+1)=(n+2)^2. an
Sn+1 = (n+2)^2. an / [4(n+1) ] (1)
S(n-1)+1 = (n+1)^2. a(n-1) / (4n) (2)
(1)-(2)
an = (n+2)^2. an / [4(n+1) ] - (n+1)^2. a(n-1) / (4n)
4n(n+1)an = n.(n+2)^2an - (n+1)^3.a(n-1)
n[ (n+2)^2-4(n+1) ] .an = (n+1)^3.a(n-1)
n^3. an = (n+1)^3.a(n-1)
an/a(n-1) = [(n+1)/n]^3
an/a1 = [(n+1)/2]^3
an = (n+1)^3