摘要:(II)设.且1<a1<2.求证+-+<2.[标准答案]
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设函数f(x)=
,且a1=
, an+1=f(an),其中n=1,2,3,….
(I)计算a2,a3的值;
(II)设a2=2,求证:数列{bn}为等比数列;
(III)求证:
≤an<1.
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2x |
x+1 |
1 |
2 |
(I)计算a2,a3的值;
(II)设a2=2,求证:数列{bn}为等比数列;
(III)求证:
1 |
2 |
已知函数f(x)=eλx+(1-λ)a-λex,其中α,λ,是常数,且0<λ<1.
(I)求函数f(x)的极值;
(II)对任意给定的正实数a,是否存在正数x,使不等式||<a成立?若存在,求出x,若不存在,说明理由;
(III)设λ1,λ2∈(0,+∞),且λ1+λ2=1,证明:对任意正数a1,a2都有:≤λ1a1+λ2a2.
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设数列{bn}满足b1=1,bn+1=2bn+1,若数列{an}满足:a1=1,且当n≥2,n∈N*时,an=bn(
+
+…+
)
(I) 求b2,b3,b4及bn;
(II)证明:
(1+
)<
(n∈N*),(注:
(1+
)=(1+
)(1+
)…(1+
)).
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1 |
b1 |
1 |
b2 |
1 |
bn-1 |
(I) 求b2,b3,b4及bn;
(II)证明:
n |
k=1 |
1 |
ak |
10 |
3 |
n |
k=1 |
1 |
ak |
1 |
a1 |
1 |
a2 |
1 |
an |