数列112+2,122+4,132+6,142+8,…前n项的和等于______.-数学

题目简介

数列112+2,122+4,132+6,142+8,…前n项的和等于______.-数学

题目详情

数列
1
12+2
1
22+4
1
32+6
1
42+8
,…
前n项的和等于______.
题型:填空题难度:中档来源:不详

答案

an=class="stub"1
n2+2n
=class="stub"1
n(n+2)
 =class="stub"1
2
(class="stub"1
n
-class="stub"1
n+2
 )

∴Sn=a1+a2+a3+…+an
=class="stub"1
2
(1-class="stub"1
3
) +class="stub"1
2
(class="stub"1
2
-class="stub"1
4
)+class="stub"1
2
( class="stub"1
3
-class="stub"1
5
)+…+class="stub"1
2
(class="stub"1
n
-class="stub"1
n+2
)

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

=class="stub"3
4
-class="stub"2n+3
2(n+1)(n+2)

故答案为:class="stub"3
4
-class="stub"2n+3
2(n+1)(n+2)

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