把正整数按“S”型排成了如图所示的三角形数表,第n行有n个数,设第n行左侧第一个数为an,如a5=15,则该数列{an}的前n项和Tn(n为偶数)为()A.Tn=n(n+1)(2n+1)10B.Tn=

题目简介

把正整数按“S”型排成了如图所示的三角形数表,第n行有n个数,设第n行左侧第一个数为an,如a5=15,则该数列{an}的前n项和Tn(n为偶数)为()A.Tn=n(n+1)(2n+1)10B.Tn=

题目详情

把正整数按“S”型排成了如图所示的三角形数表,第n行有n个数,设第n行左侧第一个数为an,如a5=15,则该数列{an}的前n项和Tn(n为偶数)为(  )
A.Tn=
n(n+1)(2n+1)
10
B.Tn=
n3
6
+
n2
4
+
n
3
C.Tn=
n3
6
+
n2
4
-
n
6
D.Tn=
n(n+1)(n+2)
6
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题型:单选题难度:偏易来源:蓝山县模拟

答案

方法一:(特值法)因为T2=a1+a2=3,把n=2代入选项,排除C、D,再代入n=4,因为T4=16,B选项满足,故选B.
方法二:因为当n为奇数时,an=1+2+…+n=
n(n+1)
2
,当n为偶数时,an=an-1+1,
故n是偶数时,Tn=a1+(a1+1)+a3+(a3+1)+…+an-1+(an-1+1)
=2a1+1+2a3+1+…+2an-1+1
=2(a1+a3+…+an-1)+class="stub"n
2

=1×2+3×4+…+(n-1)n+class="stub"n
2

=(12+1)+(32+3)+…+[(n-1)2+(n-1)]+class="stub"n
2

=[12+32+52+…+(n-1)2]+[1+3+…+(n-1)]+class="stub"n
2

令S=12+22+…+(n-1)2+n2,A=12+32+52+…+(n-1)2,B=22+42+62+…+n2,
A-B=12-22+32-42+52-62+…+(n-1)2-n2=-1-2-3-4-…-(n-1)-n=-
n(n+1)
2

A+B=
n(n+1)(2n+1)
6
,得A=
n(n+1)(2n+1)
6
-
n(n+1)
2
2
=
n(n+1)(n-1)
6

则 Tn=
n(n+1)(n-1)
6
+
(1+n-1)•class="stub"n
2
2
+class="stub"n
2
=
n(n2-1)
6
+
n2
4
+class="stub"n
2
=
n3
6
+
n2
4
+class="stub"n
3

故选B.

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