作业帮f(x)满足f(x+y)=f(x)×f(y),且f(1)=0.5.f(x)满足f(x+y)=f(x)×f(y),且f(1)=0.5.1,当n∈N,求f(n)的表达式2,设an=×f(n),n∈N,求证:a1+a2+a3+.+an＜23设bn=n×f(n+1)/f(n),n∈N,Sn为{bn}的前n项和,求1/S1+1/S2+...+1/Sn.2,设an=n×f(n),n∈N,
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f(x)满足f(x+y)=f(x)×f(y),且f(1)=0.5.f(x)满足f(x+y)=f(x)×f(y),且f(1)=0.5.1,当n∈N,求f(n)的表达式2,设an=×f(n),n∈N,求证:a1+a2+a3+.+an＜23设bn=n×f(n+1)/f(n),n∈N,Sn为{bn}的前n项和,求1/S1+1/S2+...+1/Sn.2,设an=n×f(n),n∈N,
f(x)满足f(x+y)=f(x)×f(y),且f(1)=0.5.
f(x)满足f(x+y)=f(x)×f(y),且f(1)=0.5.
1,当n∈N,求f(n)的表达式
2,设an=×f(n),n∈N,求证:a1+a2+a3+.+an＜2
3设bn=n×f(n+1)/f(n),n∈N,Sn为{bn}的前n项和,求1/S1+1/S2+...+1/Sn.
2,设an=n×f(n),n∈N,求证:a1+a2+a3+.....+an＜2
f(x)满足f(x+y)=f(x)×f(y),且f(1)=0.5.f(x)满足f(x+y)=f(x)×f(y),且f(1)=0.5.1,当n∈N,求f(n)的表达式2,设an=×f(n),n∈N,求证:a1+a2+a3+.+an＜23设bn=n×f(n+1)/f(n),n∈N,Sn为{bn}的前n项和,求1/S1+1/S2+...+1/Sn.2,设an=n×f(n),n∈N,
f(n)=f((n-1)+1)=f(n-1)*f(1)=f(n-1)*0.5=f(n-1)/2
同理 f(n+1)=f(n)/2
所以f(2)=f(1)/2 f(n)=f(1)/2(n-1)(分母为2的n-1次方 我打不出来)
f(n)=1/2n(分母为2的n次方 我打不出来)
a1+a2+……an=a1+2a2+……nan=1/2+1/4+1/8+……+1/2n=(1/2+1/2n)*2/n=(n+1)/n=1+1/n
因为n∈N 所以1/n
查查解数列的方法即可
(1)f(n+1)=f(n)*f(1)
f(n+1)=0.5f(n)
f(n)=0.5^n
(2)an=n*0.5^n
an=n/2^n
Sn=1/2^1+2/2^2+……+(n-1)/2^(n-1)+n/2^n
两边同乘1/2
1/2Sn=1/2^2+2/2^3……+（n-1)/2^n+n/2^(n+1)
错位相减<...
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(1)f(n+1)=f(n)*f(1)
f(n+1)=0.5f(n)
f(n)=0.5^n
(2)an=n*0.5^n
an=n/2^n
Sn=1/2^1+2/2^2+……+(n-1)/2^(n-1)+n/2^n
两边同乘1/2
1/2Sn=1/2^2+2/2^3……+（n-1)/2^n+n/2^(n+1)
错位相减
1/2Sn=1/2+1/2^2+1/2^3+……+1/2^n-n/2^(n+1)
1/2Sn=1-1/2^n-n/2^(n+1)
Sn=2-1/2^(n-1)-n/2^(n+1)
因为后面的2项都为正数
所以Sn<2
(3)bn=n/2
Sn=n(n+1)/4
1/Sn=4/n(n+1)
1/Sn=4[1/n-1/(n+1)]
裂项相加
1/S1+1/S2+……+1/Sn=4[1-1/(n+1)]
收起
f(n)=f(1*n)=(f(1))∧n=0.5∧n
下面有点麻烦
试试化简，查查解数列的方法即可