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已知f(n)=cosnπ/4,n属于正整数.则f(1)+f(2)+f(3)+……f(100)=多少
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已知f(n)=cosnπ/4,n属于正整数.则f(1)+f(2)+f(3)+……f(100)=多少
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答案和解析
因为f(n)=cosnπ/4
所以对于任意k为非负整数
f(8k+1)+f(8k+2)+f(8k+3)+f(8k+4)+f(8k+5)+f(8k+6)+f(8k+7)+f(8k+8)
=cos(2kπ+π/4)+cos(2kπ+2π/4)+cos(2kπ+3π/4)+cos(2kπ+4π/4)
+cos(2kπ+5π/4)+cos(2kπ+6π/4)+cos(2kπ+7π/4)+cos(2kπ+8π/4)
=cosπ/4+cos2π/4+cos3π/4+cos4π/4+cos5π/4+cos6π/4+cos7π/4+cos8π/4
=cosπ/4+cos3π/4-1+cos5π/4+cos7π/4+1
=cosπ/4+cos3π/4+cos5π/4+cos7π/4
=0+0
=0
所以f(1)+f(2)+f(3)+……+f(8)=0
f(9)+f(10)+f(11)+……+f(16)=0
……
f(89)+f(90)+f(91)+……+f(96)=0
所以f(1)+f(2)+f(3)+……f(100)=f(97)+f(98)+f(99)+f(100)
=f(8*12+1)+f(8*12+2)+f(8*12+3)+f(8*12+4)
=f(1)+f(2)+f(3)+f(4)
=cosπ/4+cos2π/4+cos3π/4+cos4π/4
=√2/2+0-√2/2-1
=-1
所以对于任意k为非负整数
f(8k+1)+f(8k+2)+f(8k+3)+f(8k+4)+f(8k+5)+f(8k+6)+f(8k+7)+f(8k+8)
=cos(2kπ+π/4)+cos(2kπ+2π/4)+cos(2kπ+3π/4)+cos(2kπ+4π/4)
+cos(2kπ+5π/4)+cos(2kπ+6π/4)+cos(2kπ+7π/4)+cos(2kπ+8π/4)
=cosπ/4+cos2π/4+cos3π/4+cos4π/4+cos5π/4+cos6π/4+cos7π/4+cos8π/4
=cosπ/4+cos3π/4-1+cos5π/4+cos7π/4+1
=cosπ/4+cos3π/4+cos5π/4+cos7π/4
=0+0
=0
所以f(1)+f(2)+f(3)+……+f(8)=0
f(9)+f(10)+f(11)+……+f(16)=0
……
f(89)+f(90)+f(91)+……+f(96)=0
所以f(1)+f(2)+f(3)+……f(100)=f(97)+f(98)+f(99)+f(100)
=f(8*12+1)+f(8*12+2)+f(8*12+3)+f(8*12+4)
=f(1)+f(2)+f(3)+f(4)
=cosπ/4+cos2π/4+cos3π/4+cos4π/4
=√2/2+0-√2/2-1
=-1
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