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根据杨辉三角形公式(a+b)的五次方是多少
人气:430 ℃ 时间:2019-12-24 20:34:46
解答
(a+b)^5
=(a+b)^2*(a+b)^3
=(a^2+2ab+b^2)*(a^3+3a^2b+3ab^2+b^3)
=a^2*(a^3+3a^2b+3ab^2+b^3)
+2ab*(a^3+3a^2b+3ab^2+b^3)
+b^2*(a^3+3a^2b+3ab^2+b^3)
=(a^5+3a^4b+3a^3b^2+a^2b^3)
+(2a^4b+6a^3b^2+6a^2b^3+2ab^4)
+(a^3b^2+3a^2b^3+3ab^4+b^5)
=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5
所以杨辉三角第6行1,5,10,10,5,1
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