数列{an}的通项公式an=nsin(n+12π)+1,前n项和为Sn(n∈N*),则S2013=()A.1232B.2580C.3019D.4321-数学

题目简介

数列{an}的通项公式an=nsin(n+12π)+1,前n项和为Sn(n∈N*),则S2013=()A.1232B.2580C.3019D.4321-数学

题目详情

数列{an}的通项公式an=nsin(
n+1
2
π
)+1,前n项和为Sn(n∈N*),则S2013=(  )
A.1232B.2580C.3019D.4321
题型:单选题难度:中档来源:保定一模

答案

当n=4k(k∈Z)时,sin(class="stub"n+1
2
π
)=sinclass="stub"π
2
=1;当n=4k+1(k∈Z)时,sin(class="stub"n+1
2
π
)=sinπ=0
当n=4k+2(k∈Z)时,sin(class="stub"n+1
2
π
)=sinclass="stub"3π
2
=-1;当n=4k+3(k∈Z)时,sin(class="stub"n+1
2
π
)=sin2π=0
由此可得
S2013=(1×sinπ+1)+(2×sinclass="stub"3π
2
+1)+(3×sin2π+1)+…+(2013sinclass="stub"2014
2
π
+1)
=[2×(-1)+4×1+6×(-1)+8×1+…+2010×(-1)+2012×1]+2013×1
=(-2+4-6+8-10+…+2008-2010+2012)+2013=1006+2013=3019
故选:C

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