(1)∵等比数列{a n}的前n项和为S n=a·2 n+b,且a 1=3.
∴a 1=2a+b=3,a 2=4a+b﹣(2a+b)=2a,a 3=(8a+b)﹣(4a+b)=4a,
∴公比q=
=2.
∵
,∴a=3,b=﹣3.
∴a n=3·2 n﹣1
(2)b n=
=
,T n=
(1+
+
+…+
)④
T n=
(
+
+…+
+
)⑤
④﹣⑤得:
T n=
(1+
+
+…+
﹣
)=
(
)
=
(2﹣
﹣
)=
(1﹣
﹣
),
∴T n=
(1﹣
﹣
).
(1)∵等比数列{a n}的前n项和为S n=a·2 n+b,且a 1=3.
∴a 1=2a+b=3,a 2=4a+b﹣(2a+b)=2a,a 3=(8a+b)﹣(4a+b)=4a,
∴公比q=
=2.
∵
,∴a=3,b=﹣3.
∴a n=3·2 n﹣1
(2)b n=
=
,T n=
(1+
+
+…+
)④
T n=
(
+
+…+
+
)⑤
④﹣⑤得:
T n=
(1+
+
+…+
﹣
)=
(
)
=
(2﹣
﹣
)=
(1﹣
﹣
),
∴T n=
(1﹣
﹣
).