Question

数列{a n } 的通项a n =n²(cos²nπ/3-sin²nπ/3),其前n项和为Sn,则S30为?

Answer 1

an =n²(cos²nπ/3-sin²nπ/3)=n²*cos(2nπ/3)
对于函数f(n)=cos(2nπ/3)，周期T=2π/(2π/3)=3
则可知：
当n=3k+1，k属于N时，an=(3k+1)²*cos[2(3k+1)π/3]
=(3k+1)²*cos(2kπ+2π/3)
=(3k+1)²*cos(2π/3)= - (3k+1)²/2
当n=3k+2，k属于N时，an=(3k+2)²*cos[2(3k+2)π/3]
=(3k+2)²*cos(2kπ+4π/3)
=(3k+2)²*cos(4π/3)= - (3k+2)²/2
当n=3k+3，k属于N时，an=(3k+3)²*cos[2(3k+3)π/3]
=(3k+3)²*cos(2kπ+2π)
=(3k+3)²
所以：
a(3k+1) + a(3k+2) + a(3k+3)
= - (3k+1)²/2 - (3k+2)²/2 + (3k+3)²
=(1/2)*[2(3k+3)²- (3k+2)² - (3k+1)²]
=(1/2)*[(18k²+36k+18)²- (9k²+12k+4) - (9k²+6k+1)]
=(1/2)*(18k+13)
=9k + 13/2
由于30÷3=10，所以：
S30=(a1+a2+a3)+(a4+a5+a6)+...+(a28+a29+a30)
=9*(0+1+2+3+...+9)+(13/2)*10
=9*45+65
=405+65
=470

Answer 2

这个数列有周期，你多写几项，规律就出来了

追问

但是我求出的答案是几千（这是选择题给出的都是几百），我不知道哪里理解错了，还是算错了