证明:(1)a
n=(1+
)
n=1+C
n1+C
n2(
)
2+…+C
nn(
)
n,a
n+1=(1+
)
n+1=1+C
n+11+C
n+12(
)
2+…+C
n+1n(
)
n+1.
可观察C
n+1k(
)
k与C
nk(
)
k,当k=0,1时,
C
n+1k(
)
k=C
nk(
)
k;当k=2,3,4,,n时,
C
n+1k(
)
k>C
nk(
)
k.∴a
n<a
n+1,即{a
n}为递增数列.
(2)∵a
n=(1+
)
n=1+C
n1+C
n2(
)
2++C
nn(
)
n≥1+C
n1=2,
又a
n=(1+
)
n=1+C
n1+C
n2(
)
2++C
nn(
)
n≤2+
+
+…+
=3-
<3.
分析:(1)由题设条件知a
n=1+C
n1+C
n2(
)
2+…+C
nn(
)
n,a
n+1=(1+
)
n+1=1+C
n+11+C
n+12(
)
2+…+C
n+1n(
)
n+1.由此可知a
n<a
n+1,即{a
n}为递增数列.
(2)由题意知a
n=1+C
n1+C
n2(
)
2++C
nn(
)
n≥1+C
n1=2,由此可知a
n=1+C
n1+C
n2(
)
2++C
nn(
)
n≤2+
+
+…+
=3-
<3.
点评:解:本题考查数列的综合应用,解题时要注意公式的灵活运用.