数列{a(n)}的通项a(n)=n²(cos²nπ/3- sin²nπ/3),其前n项和为Sn,则S(30)=?
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![数列{a(n)}的通项a(n)=n²(cos²nπ/3- sin²nπ/3),其前n项和为Sn,则S(30)=?](/uploads/image/z/5086433-65-3.jpg?t=%E6%95%B0%E5%88%97%7Ba%28n%29%7D%E7%9A%84%E9%80%9A%E9%A1%B9a%28n%29%3Dn%26%23178%3B%28cos%26%23178%3Bn%CF%80%2F3-+sin%26%23178%3Bn%CF%80%2F3%29%2C%E5%85%B6%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%88%99S%2830%29%3D%3F)
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数列{a(n)}的通项a(n)=n²(cos²nπ/3- sin²nπ/3),其前n项和为Sn,则S(30)=?
数列{a(n)}的通项a(n)=n²(cos²nπ/3- sin²nπ/3),其前n项和为Sn,则S(30)=?
数列{a(n)}的通项a(n)=n²(cos²nπ/3- sin²nπ/3),其前n项和为Sn,则S(30)=?
COS²nπ/3 无论n如何变化都为 1/4
SIN²nπ/3 无论n如何变化都为 3/4
∴ a(n)= - 1/2 *n²
s(n) = - 1/2 *n²*(1²+2²+.+30²)
平方和公式 :1²+2²+.+n² =n*(n+1)*(2n+1)/6
然后你自己处理吧 .嘿嘿
cos²nπ/3- sin²nπ/3=cos2nπ/3
所以S(30)=前30项平方和/3