考點(diǎn):數(shù)列的求和,等差關(guān)系的確定
專題:等差數(shù)列與等比數(shù)列
分析:(1)由已知條件得S
n=
(an+1)2,從而得得a
1=1,a
n-a
n-1=2,由此能證明數(shù)列{a
n}是首項(xiàng)為1,公差為2的等差數(shù)列.
(2)由已知條件得b
1=4,b
n+3=2(b
n-1+3),由此求出b
n=2
n+1-3.
(3)c
n=
=
,由此利用裂項(xiàng)求和法能求出數(shù)列{c
n}的前n項(xiàng)和T
n.
解答:
(1)證明:正項(xiàng)數(shù)列{a
n}的前n項(xiàng)和為S
n,
是
與(a
n+1)
2的等比中項(xiàng),
∴S
n=
(an+1)2,
∴a
1=S
1=
(a
1+1)
2,
解得a
1=1,
n≥2時(shí),a
n=
(a
n+1)
2-
(a
n-1+1)
2,
整理,得(a
n+a
n+1)[
(a
n-a
n-1)-
]=0,
∵a
n>0,∴
(a
n-a
n-1)-
=0,
∴a
n-a
n-1=2,
∴數(shù)列{a
n}是首項(xiàng)為1,公差為2的等差數(shù)列.
(2)解:∵b
1=a
1,且b
n=2b
n-1+3(n≥2),
∴b
1=1,b
n+3=2(b
n-1+3),
∴{b
n+3}是首項(xiàng)為4,公比為2的等比數(shù)列,
∴b
n+3=4•2
n-1=2
n+1,∴b
n=2
n+1-3.
(3)解:∵數(shù)列{a
n}是首項(xiàng)為1,公差為2的等差數(shù)列,
∴a
n=1+(n-1)×2=2n-1,
∴c
n=
=
,
∴
Tn=+++…+,①
Tn=
+++…+,②
①-②,得:
Tn=
+2(
+
+…+
)-
=
+2×
-
=
+1-
-
=
-
.
∴T
n=3-
.
點(diǎn)評:本題考查等差數(shù)列的證明,考查數(shù)列的通項(xiàng)公式和前n項(xiàng)和公式的求法,解題時(shí)要認(rèn)真審題,注意錯位相減法的合理運(yùn)用.