(1-x+x的平方)(1+x)的八次方展开式+x的四次方

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摘要总之正确答案是∵(a+b)^7=a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7∴(x+1)^7=x^7+7x^6+21x^5+35x^4+35x^3+21x^2+7x+1∴(1-x+x^2)(1+x)^8=(x^3+1)(x+1)^7=(x^3+1)(x^7+7x^6+21x^5+35x^4+35x^3+21x^2+7x+1)=x^10+7x^9+21x^8+35x^7+35x^6+21x^5+7x^4+x^3+x^7+7x^6+21x^5+35x^4+35x^3+21x^2+7x+1=x^10+7x^9+21x^8+36x^7+42x^6+42x^5+42x^4+36x^3+21x^2+7x+1∴(1-x+x的平方)(1+x)的八次方展开式中x^4的系数是42

咨询记录 · 回答于2023-04-20

(1-x+x的平方)(1+x)的八次方展开式+x的四次方

您问的是(1-x+x的平方)(1+x)的八次方展开式中x的四次方的系数吗？

是的

(a+b)^7=a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7(1-x+x^2)(1+x)^8=(x^3+1)(x+1)^7=(x^3+1)(x^7+7x^6+21x^5+35x^4+35x^3+21x^2+7x+1)=x^10+7x^9+21x^8+35x^7+35x^6+21x^5+7x^4+x^3+x^7+7x^6+21x^5+35x^4+35x^3+21x^2+7x+1x^4的系数是42

总之正确答案是∵(a+b)^7=a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7∴(x+1)^7=x^7+7x^6+21x^5+35x^4+35x^3+21x^2+7x+1∴(1-x+x^2)(1+x)^8=(x^3+1)(x+1)^7=(x^3+1)(x^7+7x^6+21x^5+35x^4+35x^3+21x^2+7x+1)=x^10+7x^9+21x^8+35x^7+35x^6+21x^5+7x^4+x^3+x^7+7x^6+21x^5+35x^4+35x^3+21x^2+7x+1=x^10+7x^9+21x^8+36x^7+42x^6+42x^5+42x^4+36x^3+21x^2+7x+1∴(1-x+x的平方)(1+x)的八次方展开式中x^4的系数是42

您问的是我回答内容的意思吗？

已知无穷数列an满足a1=1 a2=0 a3=

-1

已知无穷数列an满足a1=1 a2=0 a3=-1

已知无穷数列an满足a1=1 a2=0 a3=-1 a4=0 写一个通项公式

您好，无穷数列an满足a1=1 a2=0 a3=-1 a4=0 写一个通项公式答，an=sin(nπ/2)或an=cos(nπ/2-π/2)(n∈N)

N是自然数集。

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