已知函数f(x)=x2-2x+4,数列{an}是公差为d的等差数列,若a1=f(d-1),a3=f(d+1)(1)求数列{an}的通项公式;(2)Sn为{an}的前n项和,求证:1S1+1S2+…+1

题目简介

已知函数f(x)=x2-2x+4,数列{an}是公差为d的等差数列,若a1=f(d-1),a3=f(d+1)(1)求数列{an}的通项公式;(2)Sn为{an}的前n项和,求证:1S1+1S2+…+1

题目详情

已知函数f(x)=x2-2x+4,数列{an}是公差为d的等差数列,若a1=f(d-1),a3=f(d+1)
(1)求数列{an}的通项公式;
(2)Sn为{an}的前n项和,求证:
1
S1
+
1
S2
+…+
1
Sn
1
3
题型:解答题难度:中档来源:不详

答案

(1)a1=f(d-1)=d2-4d+7,a3=f(d+1)=d2+3,
又由a3=a1+2d,可得d=2,所以a1=3,an=2n+1
(2)证明:由题意,Sn=
n(3+2n+1)
2
=n(n+2)

所以,class="stub"1
Sn
=class="stub"1
n(n+2)
=class="stub"1
2
(class="stub"1
n
-class="stub"1
n+2
)

所以,class="stub"1
S1
+class="stub"1
S2
+…+class="stub"1
Sn
=class="stub"1
2
(1-class="stub"1
3
+class="stub"1
2
-class="stub"1
4
+class="stub"1
3
-class="stub"1
5
+…+class="stub"1
n
-class="stub"1
n+2
)

=class="stub"1
2
(class="stub"3
2
-class="stub"1
n+1
-class="stub"1
n+2
)
class="stub"1
2
(class="stub"3
2
-class="stub"1
1+1
-class="stub"1
1+2
)
=class="stub"1
3

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