题目内容
设{an}是一个公差为d(d≠0)的等差数列,它的前n项和为Sn,S10=110且a1,a2,a4成等比数列.
(Ⅰ)证明a1=d;
(Ⅱ)求公差d的值和数列{an}的前n项和Sn;
(Ⅲ)设bn=
,求数列{bn}的前n项和Tn.
(Ⅰ)证明a1=d;
(Ⅱ)求公差d的值和数列{an}的前n项和Sn;
(Ⅲ)设bn=
1 | Sn |
分析:(Ⅰ)由a1,a2,a4成等比数列,可得
=a1a4,再由{an}是等差数列,可得a2=a1+d,a4=a1+3d,代入化简可得;(Ⅱ)由等差数列的求和公式代入已知条件可得d的值,进而可得a1的值,可得通项公式,进而可得前n项和;( III)可得bn=
=
=
-
,裂项相消法可得其和.
a | 2 2 |
1 |
Sn |
1 |
n(n+1) |
1 |
n |
1 |
n+1 |
解答:解:(Ⅰ)因a1,a2,a4成等比数列,故
=a1a4,
又∵{an}是等差数列,有a2=a1+d,a4=a1+3d,
∴(a1+d)2=a1(a1+3d),
+2a1d+d2=
+3a1d,
化简可得d2=a1d,又∵d≠0,解得a1=d
(Ⅱ)由等差数列的求和公式可得S10=10a1+
d,
化简可得10a1+45d=110,把a1=d代入上式得55d=110,
解得d=2,∴a1=2,∴an=a1+(n-1)d=2n.
∴Sn=
=n2+n
( III)由(Ⅱ)得bn=
=
=
-
,
∴Tn=b1+b2+…+bn=(1-
)+(
-
)+…+(
-
)
=1-
=
,即Tn=
a | 2 2 |
又∵{an}是等差数列,有a2=a1+d,a4=a1+3d,
∴(a1+d)2=a1(a1+3d),
a | 2 1 |
a | 2 1 |
化简可得d2=a1d,又∵d≠0,解得a1=d
(Ⅱ)由等差数列的求和公式可得S10=10a1+
10×9 |
2 |
化简可得10a1+45d=110,把a1=d代入上式得55d=110,
解得d=2,∴a1=2,∴an=a1+(n-1)d=2n.
∴Sn=
n(2+2n) |
2 |
( III)由(Ⅱ)得bn=
1 |
Sn |
1 |
n(n+1) |
1 |
n |
1 |
n+1 |
∴Tn=b1+b2+…+bn=(1-
1 |
2 |
1 |
2 |
1 |
3 |
1 |
n |
1 |
n+1 |
=1-
1 |
n+1 |
n |
n+1 |
n |
n+1 |
点评:本题考查等差数列的通项公式和求和公式,涉及裂项相消法求数列的和,属中档题.

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