设函数f(x-1)=x+x 2 +x 3 +…+x n (x≠0,1),且f(x)中所有项的系数和为a n ,则 lim n→∞

2025-05-15 23:08:59
推荐回答(1个)
回答1:

函数f(x-1)=x+x 2 +x 3 +…+x n =
x(1- x n )
1-x
,所以f(x)=
(x+1)[ (x+1) n -1]
x

当x=1时,f(x)中所有项的系数和为a n =2×(2 n -1)=2×2 n -2,
lim
n→∞
a n
2 n
=
lim
n→∞
2×2 n  -2
2 n
=2-
lim
n→∞
2
2 n
=2.
故答案为:2.