问题
解答题
已知函数f(x)=
(1)证明:对∀x∈[1,+∞),f(x)<g(x)恒成立; (2)n∈N*时,证明:
|
答案
证明:(1)∵f(x)=
,外函数y=3x+1 3x+1-1
是减函数,内函数t=3x是增函数t+1 3t-1
∴f(x)在R上递减
∵g(x)=
在[1,+∞)上是增函数3x x+1
∴f(x)-g(x)在[1,+∞)是减函数
∴f(x)-g(x)≤f(1)-g(1)=-1<0
(2)
+n 3n+1
<n+1 3n+1-1
+n 3n
⇔n+1 3n+1
-n 3n+1
<n 3n
-n+1 3n+1
⇔n+1 3n+1-1
<-n 3n+1
⇔-(n+1) 3(3n+1-1)
<3n+1 3n+1-1
已证3n n+1
∴
+n 3n+(-1)n-1
<n+1 3n+1+(-1)n
+n 3n
(n为奇数时)n+1 3n+1
∴当n为奇数时,
+1 3+1
+…+2 32-1
+n 3n+1
<(n+1 3n+1-1
+1 3
)+…+(2 32
+n 3n
)n+1 3n+1
由错位相减法可得:
+1 3
+…+2 32
=n+1 3n+1
-3 4
-1 4 • 3n
<n+1 2 • 3n+1 3 4
当n为偶数时,所求
+1 3+1
+…+2 32-1
+n 3n-1
<n+1 3n+1+1
+…+1 3+1
+n+1 3n+1+1
<n+2 3n+2-1 3 4
綜上,原不等式成立,即
+1 3+1
+2 32-1
+…+3 33+1
+n 3n+(-1)n-1
<n+1 3n+1+(-1)n 3 4