已知數(shù)列{an}滿足a1=2a,an=2a-a2an?1(n≥2),其中a是不為0的常數(shù),令bn=1/an?a. (1)求證:數(shù)列{bn}是等差數(shù)列; (2)求數(shù)列{an}的通項(xiàng)公式.
已知數(shù)列{a
n}滿足a
1=2a,a
n=2a-
(n≥2),其中a是不為0的常數(shù),令b
n=
.
(1)求證:數(shù)列{b
n}是等差數(shù)列;
(2)求數(shù)列{a
n}的通項(xiàng)公式.
優(yōu)質(zhì)解答
∵(1)a
n=2a-
(n≥2),
∴b
n=
==(n≥2),
∴b
n-b
n-1=
?=(n≥2),
∴數(shù)列{b
n}是公差為
的等差數(shù)列.
(2)∵b
1=
=
,
故由(1)得:b
n=
+(n-1)×
=
.
即:
=
,
得:a
n=a(1+
).
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