S=(cos^2 x+cos^4 x+.+cos^2n x)+(sin^2x+sin^4 x+.+sin^2n x)数列
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S=(cos^2 x+cos^4 x+.+cos^2n x)+(sin^2x+sin^4 x+.+sin^2n x)数列求和题!
答案是(cos^4 x+sin^4 x-cos^2n+4 x-sin^2n+4 x)/sin^2 x.cos^2 x
答案是(cos^4 x+sin^4 x-cos^2n+4 x-sin^2n+4 x)/sin^2 x.cos^2 x
提示:两个等比数列和,再求和.
S=(cos²x)[1-(cos²x)ⁿ]/(1-cos²x) +(sin²x)[1-(sin²x)ⁿ]/(1-sin²x)
=(cos²x)[1-(cos²x)ⁿ]/sin²x +(sin²x)[1-(sin²x)ⁿ]/cos²x
=[(cos⁴x)-(cosx)^(2n+4)+(sin⁴x)-(sinx)^(2n+4)]/(sin²xcos²x)
S=(cos²x)[1-(cos²x)ⁿ]/(1-cos²x) +(sin²x)[1-(sin²x)ⁿ]/(1-sin²x)
=(cos²x)[1-(cos²x)ⁿ]/sin²x +(sin²x)[1-(sin²x)ⁿ]/cos²x
=[(cos⁴x)-(cosx)^(2n+4)+(sin⁴x)-(sinx)^(2n+4)]/(sin²xcos²x)
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