(1)见解析(2)
-
(1)证明 ∵a n=2-
,∴a n +1=2-
.
∴b n +1-b n=
-
=
-
=
=1,
∴{b n}是首项为b 1=
=1,公差为1的等差数列.
(2)解 由(1)知b n=n,
∴c n=
=
=
(
-
),
∴S n=
[(1-
)+(
-
)+(
-
)+…+(
-
)+(
-
)]
=
(1+
-
-
)=
-
.
(1)见解析(2)
-
(1)证明 ∵a n=2-
,∴a n +1=2-
.
∴b n +1-b n=
-
=
-
=
=1,
∴{b n}是首项为b 1=
=1,公差为1的等差数列.
(2)解 由(1)知b n=n,
∴c n=
=
=
(
-
),
∴S n=
[(1-
)+(
-
)+(
-
)+…+(
-
)+(
-
)]
=
(1+
-
-
)=
-
.