考點(diǎn):數(shù)列的求和,數(shù)列遞推式
專題:綜合題,等差數(shù)列與等比數(shù)列
分析:(1)利用
=a
n+1-
n
2-n-
,代入計(jì)算,即可求a
2的值;
(2)再寫一式,兩式相減,即可求數(shù)列{a
n}的通項(xiàng)公式;
(3)分類討論,證明當(dāng)n≥3時(shí),n
2>(n-1)•(n+1),可得
<
,利用裂項(xiàng)法求和,可得結(jié)論.
解答:
(1)解:∵
=a
n+1-
n
2-n-
,n∈N?.
∴當(dāng)n=1時(shí),2a
1=2S
1=a
2-
-1-
=a
2-2.
又a
1=1,∴a
2=4.
(2)解:∵
=a
n+1-
n
2-n-
,n∈N?.
∴2S
n=na
n+1-
n
3-n
2-
n
=na
n+1-
,①
∴當(dāng)n≥2時(shí),2S
n-1=(n-1)a
n-
,②
由①-②,得2S
n-2S
n-1=na
n+1-(n-1)a
n-n(n+1),
∵2a
n=2S
n-2S
n-1,∴2a
n=na
n+1-(n-1)a
n-n(n+1),
∴
-
=1,∴數(shù)列{a
n}是以首項(xiàng)為1,公差為1的等差數(shù)列.
∴
=1+1×(n-1)=n,∴a
n=n
2(n≥2),
當(dāng)n=1時(shí),上式顯然成立.∴a
n=n
2,n∈N
*.
(3)證明:由(2)知,a
n=n
2,n∈N
*,
①當(dāng)n=1時(shí),
=1<
,∴原不等式成立.
②當(dāng)n=2時(shí),
+
=1+
<
,∴原不等式成立.
③當(dāng)n≥3時(shí),∵n
2>(n-1)•(n+1),
∴
<
,
∴
+
+…+
<1+
+
+…+
+
=1+
(
-
+
-
+
-
+…+
-
+
-
)
=1+
(
-
-
)<
,
∴當(dāng)n≥3時(shí),∴原不等式亦成立.
綜上,對一切正整數(shù)n,有
+
+…+
<
.
點(diǎn)評:本題考查數(shù)列遞推式,考查數(shù)列求和,考查裂項(xiàng)法的運(yùn)用,確定數(shù)列的通項(xiàng)是關(guān)鍵.