已知函数f(x)满足f(x+y)=f(x)•f(y),且f(1)=12.(1)当x∈N+时,求f(n)的表达式;(2)设an=nf(n)&(n∈N+,求证:a1+a2+…+an<2;(3)设b

题目简介

已知函数f(x)满足f(x+y)=f(x)•f(y),且f(1)=12.(1)当x∈N+时,求f(n)的表达式;(2)设an=nf(n)&(n∈N+,求证:a1+a2+…+an<2;(3)设b

题目详情

已知函数f(x)满足f(x+y)=f(x)•f(y),且f(1)=
1
2

(1)当x∈N+时,求f(n)的表达式;
(2)设an=nf(n)
 &(n∈N+
,求证:a1+a2+…+an<2;
(3)设bn=
nf(n+1)
f(n)
 &(n∈N+),Sn=b1
+b2+…+bn
,求Sn
题型:解答题难度:中档来源:不详

答案

(1)由题设得:f(n+1)=f(n)•f(1)=class="stub"1
2
f(n)

∴数列{f(n)}是以 f(1)=class="stub"1
2
为首项,class="stub"1
2
为公比的等比数列.
f(n)=class="stub"1
2
×(class="stub"1
2
)n-1=(class="stub"1
2
)n
.(4分)

(2)设Tn=a1+a2+…+an
an=n•f(n)=n•(class="stub"1
2
)n
(n∈N*).
Tn=1×class="stub"1
2
+2×(class="stub"1
2
)2+3×(class="stub"1
2
)3++n×(class="stub"1
2
)n
class="stub"1
2
Tn

=1×(class="stub"1
2
)2+2×(class="stub"1
2
)3++(n-1)×(class="stub"1
2
)n+n×(class="stub"1
2
)n+1

两式相减得:class="stub"1
2
Tn=class="stub"1
2
+(class="stub"1
2
)2+(class="stub"1
2
)3++(class="stub"1
2
)n-n×(class="stub"1
2
)n+1

=
class="stub"1
2
×(1-class="stub"1
2n
)
1-class="stub"1
2
-n×(class="stub"1
2
)n+1=1-class="stub"n+2
2n+1

Tn=2-class="stub"n+2
2n
<2
.(10分)
(3)∵bn=
nf(n+1)
f(n)
=
n•(class="stub"1
2
)
n+1
(class="stub"1
2
)
n
=class="stub"1
2
n

Sn=class="stub"1
2
(1+2+3+…+n)

=class="stub"1
2
×class="stub"n
2
(n+1)

=
n(n+1)
4

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