问题
填空题
数列{
|
答案
∵an=
=1 4n2-1
=1 (2n-1)•(2n+1)
(1 2
-1 2n-1
)1 2n+1
∴Sn=a1+a2+…+an
=
[(1-1 2
)+(1 3
-1 3
)+…+(1 5
-1 2n-1
)]1 2n+1
=
(1-1 2
)1 2n+1
∴
Sn=lim n→∞ lim n→∞
(1-1 2
)=1 2n+1
．1 2
故答案为：1 2