已知数列{an}的通项为an=（n+2）/n*（n+1）*2^(n－1)，则其前n项和Sn=？
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an=（n+2）/[n*（n+1）*2^(n－1)]
let
(n+2)/[n(n+1)] = A/n + B/(n+1)
=> n+2 = A(n+1) +Bn
put n=0
A=2
put n=-1
B= -1
ie
an = (2/n -1/(n+1) ) (1/2)^(n-1)
Sn = a1+a2 +...+an
= (2/1 - 1/2) + ( 1/2 -1/6) + (1/6 - 1/16) +....+ { (2/n)(1/2)^(n-1) -[1/(n+1)](1/2)^(n-1) }
= 2 - 1/[(n+1). 2^(n-1)]
let
(n+2)/[n(n+1)] = A/n + B/(n+1)
=> n+2 = A(n+1) +Bn
put n=0
A=2
put n=-1
B= -1
ie
an = (2/n -1/(n+1) ) (1/2)^(n-1)
Sn = a1+a2 +...+an
= (2/1 - 1/2) + ( 1/2 -1/6) + (1/6 - 1/16) +....+ { (2/n)(1/2)^(n-1) -[1/(n+1)](1/2)^(n-1) }
= 2 - 1/[(n+1). 2^(n-1)]
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