a | n |
n(an−a1) |
2 |
a | n |
Sn+2 |
Sn+1 |
Sn+1 |
Sn+2 |
管中邪 幼苗
共回答了27个问题采纳率:85.2% 举报
nan |
2 |
(n+1)an+1 |
2 |
(Ⅰ)由已知,得S1=
1•(a−a)
2=a1=a,∴a=0….(2分)
由a1=0得Sn=
nan
2,则Sn+1=
(n+1)an+1
2,
∴2(Sn+1-Sn)=(n+1)an+1-nan,即2an+1=(n+1)an+1-nan,
于是有(n-1)an+1=nan,并且nan+2=(n+1)an+1,
∴nan+2-(n-1)an+1=(n+1)an+1-nan,即n(an+2-an+1)=n(an+1-an)
则有an+2-an+1=an+1-an,
∴{an}为等差数列;….(7分)
(Ⅱ)∵Sn=
n(n−1)p
2,∴Pn=
(n+2)(n+1)p
2
(n+1)np
2+
(n+1)np
2
(n+2)(n+1)p
2=2+
2
n−
2
n+2
∴P1+P2+P3+…+Pn−2n=(2+
2
1−
2
3)+(2+
2
2−
2
4)+…+(2+
2
n−
2
n+2)−2n=2+1−
2
n+1−
2
n+2;由n是整数可得P1+P2+P3+…+Pn-2n<3,
故存在最小的正整数M=3,使不等式P1+P2+P3+…+Pn-2n≤M恒成立….(12分)
点评:
本题考点: 数列与不等式的综合;等差关系的确定.
考点点评: 本题考查数列递推式,考查等差数列的证明,考查裂项法求数列的和,正确运用数列递推式是关键.
1年前
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