题目内容
已知数列{an}满足:a1=a+2(a≥0),an+1=
,n∈N*.
(1)若a=0,求数列{an}的通项公式;
(2)设bn=|an+1-an|,数列的前n项和为Sn,证明:Sn<a1.
an+a |
(1)若a=0,求数列{an}的通项公式;
(2)设bn=|an+1-an|,数列的前n项和为Sn,证明:Sn<a1.
(1)若a=0时,a1=2,an+1=
,
∴an+12=an且an>0.
两边取对数,得2lgan+1=lgan,
∵lga1=lg2,
∴数列{lgan}是以lg2为首项,
为公比的等比数列,
∴lgan=(
)n-1lg2,即an=221-n;
(2)由an+1=
,得an+12=an+a,①
当n≥2时,
=an-1+a,②
①-②,得(an+1+an)(an+1-an)=an-an-1,
由已知可得an>0,∴an+1-an与an-an-1同号,
∵a2=
,且a>0,∴
-
=(a+2)2-(2a+2)=a2+2a+2>0恒成立,
∴a2-a1<0,则an+1-an<0.
∵bn=|an+1-an|,∴bn=-(an+1-an),
∴Sn=-[(a2-a1)+(a3-a2)+…+(an+1-an)]=-(an+1-a1)=a1-an+1<a1.
an |
∴an+12=an且an>0.
两边取对数,得2lgan+1=lgan,
∵lga1=lg2,
∴数列{lgan}是以lg2为首项,
1 |
2 |
∴lgan=(
1 |
2 |
(2)由an+1=
an+a |
当n≥2时,
a | 2n |
①-②,得(an+1+an)(an+1-an)=an-an-1,
由已知可得an>0,∴an+1-an与an-an-1同号,
∵a2=
2a+2 |
a | 21 |
a | 22 |
∴a2-a1<0,则an+1-an<0.
∵bn=|an+1-an|,∴bn=-(an+1-an),
∴Sn=-[(a2-a1)+(a3-a2)+…+(an+1-an)]=-(an+1-a1)=a1-an+1<a1.
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