,将函数1/(1+x+x^2)展开为麦克劳林级数
f(x)=1/(1+x+x²)=(1-x)/(1-x³)
=(1-x)[1+x³+(x³)²+……+(x³)ⁿ+……]
=1-x+x³-x^4+x^6-x^7+……+x^(3n)-x^(3n+1)+……
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f(x)=1/(1+x+x²)=(1-x)/(1-x³)
=(1-x)[1+x³+(x³)²+……+(x³)ⁿ+……]
=1-x+x³-x^4+x^6-x^7+……+x^(3n)-x^(3n+1)+……