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1s1+1s2+…+1sn.-高二数学

题目简介

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1s1+1s2+…+1sn.-高二数学

题目详情

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
s1
+
1
s2
+…+
1
sn
题型:解答题难度:中档来源:不详

答案

(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+
k(k-1)
2
•d

得,2k+
k(k-1)
2
×2=2550
,解得:k=50,k=-51(舍去),
∴a=2,k=50;
(2)由sn=n•a1+
n(n-1)
2
•d
,得sn=2n+
n(n-1)
2
×2
=n(n+1),
class="stub"1
sn
=class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

class="stub"1
s1
+class="stub"1
s2
+…+class="stub"1
sn

=class="stub"1
1×2
+class="stub"1
2×3
+…+class="stub"1
n(n+1)

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

=1-class="stub"1
n+1

=class="stub"n
n+1

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