题目内容
已知等差数列前三项为a,4,3a,前n项的和为sn,sk=2550.
(1)求a及k的值;
(2)求
+
+…+
.
(1)求a及k的值;
(2)求
1 |
s1 |
1 |
s2 |
1 |
sn |
(1)设该等差数列为{an},则a1=a,a2=4,a3=3a,
由已知有a+3a=2×4,解得a1=a=2,公差d=a2-a1=4-2=2,
将sk=2550代入公式sk=ka1+
•d,
得,2k+
×2=2550,解得:k=50,k=-51(舍去),
∴a=2,k=50;
(2)由sn=n•a1+
•d,得sn=2n+
×2=n(n+1),
∴
=
=
-
,
则
+
+…+
=
+
+…+
=(1-
)+(
-
)+…+(
-
)
=1-
=
.
由已知有a+3a=2×4,解得a1=a=2,公差d=a2-a1=4-2=2,
将sk=2550代入公式sk=ka1+
k(k-1) |
2 |
得,2k+
k(k-1) |
2 |
∴a=2,k=50;
(2)由sn=n•a1+
n(n-1) |
2 |
n(n-1) |
2 |
∴
1 |
sn |
1 |
n(n+1) |
1 |
n |
1 |
n+1 |
则
1 |
s1 |
1 |
s2 |
1 |
sn |
=
1 |
1×2 |
1 |
2×3 |
1 |
n(n+1) |
=(1-
1 |
2 |
1 |
2 |
1 |
3 |
1 |
n |
1 |
n+1 |
=1-
1 |
n+1 |
=
n |
n+1 |
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