Question

求数列的和！！！数学高手来！！！在线等！！！！！

求 sinπ/6+sin2π+。。。。。+sin nπ/6 求详细解答！！！
第二个是 sin2π/6

Answer 1

设Sn=sinπ/6+sin2π/6+。。。。。+sin nπ/6，两边同乘以2*sin（π/12）
则2*（sinπ/12）*an=1/2*（sinπ/12）*sin nπ/6积化和差得cos[（2n-1）π/12]-cos[（2n+1）π/12]
因为上式第一项与a（n-1）项的第二项可以消除，故得2*sin（π/12）*Sn=cos（π/12）-cos[（2n+1）π/12]，Sn=[cos（π/12）-cos[（2n+1）π/12]]/2sin（π/12)。其他人写的答案都好奇怪哦，我只是觉得这个本来就是一个积化和差的公式运用嘛，没有必要弄得那么复杂！

Answer 2

n=1，sinπ/6=1/2
n=2，sinπ/6+sin2π/6=1/2+√3/2
n=3，sinπ/6+sin2π/6+sin3π/6=1/2+√3/2+1
n=4，sinπ/6+sin2π/6+sin3π/6+sin4π/6=1/2+√3/2+1+√3/2
n=5，sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6=1/2+√3/2+1+√3/2+1/2
n=6，sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sinπ=1/2+√3/2+1+√3/2+1/2+0
n=7，sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sinπ+sin7π/6=1/2+√3/2+1+√3/2+1/2+0-1/2
n=8，sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sinπ+sin7π/6+sin8π/6=1/2+√3/2+1+√3/2+1/2+0-1/2-√3/2
n=9，sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sinπ+sin7π/6+sin8π/6+sin9π/6=1/2+√3/2+1+√3/2+1/2+0-1/2-√3/2-1
n=10，sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sinπ+sin7π/6+sin8π/6+sin9π/6+sin10π/6=1/2+√3/2+1+√3/2+1/2+0-1/2-√3/2-1-√3/2
n=11，sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sinπ+sin7π/6+sin8π/6+sin9π/6+sin10π/6+sin11π/6=1/2+√3/2+1+√3/2+1/2+0-1/2-√3/2-1-√3/2-1/2
n=12，sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sinπ+sin7π/6+sin8π/6+sin9π/6+sin10π/6+sin11π/6+sin12π/6=1/2+√3/2+1+√3/2+1/2+0-1/2-√3/2-1-√3/2-1/2-0
n=13，sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sinπ+sin7π/6+sin8π/6+sin9π/6+sin10π/6+sin11π/6+sin12π/6+sin13π/6=1/2+√3/2+1+√3/2+1/2+0-1/2-√3/2-1-√3/2-1/2-0+1/2
以后开始新一轮的循环，所以
sinπ/6+sin2π+。。。。。+sin nπ/6
=1/2 当 n=12k+1
=(√3+1)/2 当 n=12k+2
=(√3+3)/2 当 n=12k+3
=(2√3+3)/2 当 n=12k+4
=2+√3 当 n=12k+5
=2+√3 当 n=12k+6
=(2√3+3)/2 当 n=12k+7
=(√3+3)/2 当 n=12k+8
=(√3+1)/2 当 n=12k+9
=1/2 当 n=12k+10
=0 当 n=12k+11

Answer 3

先找出sin nπ/6的小周期12
n=12k（k为正整数）时 =0
n=12k+1（k为非负整数）时 =sinπ/6=0.5
n=12k+2（k为非负整数）时 =sinπ/6+sin2π/6=0.5+0.5根号三
n=12k+3（k为非负整数）时 =sinπ/6+sin2π/6+sin3π/6=1.5+0.5根号三
n=12k+4（k为非负整数）时 =sinπ/6+sin2π/6+sin3π/6+sin4π/6=1.5+根号三
n=12k+5（k为非负整数）时 =sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6=2+根号三
n=12k+6（k为非负整数）时 =sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sin6π/6=2+根号三
n=12k+7（k为非负整数）时 =sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sin6π/6+sin7π/6=1.5+根号三
n=12k+8（k为非负整数）时 =sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sin6π/6+sin7π/6+sin8π/6=1.5+0.5根号三
n=12k+9（k为非负整数）时 =sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sin6π/6+sin7π/6+sin8π/6+sin9π/6=0.5+0.5根号三
n=12k+10（k为非负整数）时 =sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sin6π/6+sin7π/6+sin8π/6+sin9π/6+sin10π/6=0.5
n=12k+11（k为非负整数）时 =sinπ/6+sin2π/6+sin3π/6+sin4π/6+sin5π/6+sin6π/6+sin7π/6+sin8π/6+sin9π/6+sin10π/6+sin11π/6=0

Answer 4

首先构造复数 z

最后把z,z^(n+1)的值代入，等式左边复数的虚部等于等式右边复数的虚部，从而可得出通项公式

Answer 5

分类讨论，从第一个开始，每12个的和为0，