早教吧作业答案频道 -->数学-->
已知(x+2)9=a0+a1x+a2x2+……+a9x9,则(a1+3a3+5a5+7a7+9a9)2-(2a2+4a4+6a6+8a8)2的值是?
题目详情
已知(x+2)9=a0+a1x+a2x2+……+a9x9,则(a1+3a3+5a5+7a7+9a9)2-(2a2+4a4+6a6+8a8)2的值是?
▼优质解答
答案和解析
(x+2)^9=a0+a1x+a2x^2+…+a9x^9
两边求导得
9(x+2)^8=a1+2a2x+3a3x^2…+9a9x^8
令x=1,
9(1+2)^8=a1+2a2+3a3…+9a9
9*3^8=a1+2a2+3a3…+9a9
3^2*3^8=a1+2a2+3a3…+9a9
3^10=a1+2a2+3a3…+9a9
a1+2a2+3a3…+9a9=3^10
令x=-1,
9(x+2)^8=a1+2a2x+3a3x^2…+9a9x^8
9(-1+2)^8=a1-2a2+3a3…+9a9
9*1^8=a1-2a2+3a3-4a4…+9a9
9=a1-2a2+3a3-4a4…+9a9
a1-2a2+3a3-4a4…+9a9=9
(a1+3a3+5a5+7a7+9a9)^2-(2a2+4a4+6a6+8a 8)^2
=[(a1+3a3+5a5+7a7+9a9)-(2a2+4a4+6a6+8a 8)][(a1+3a3+5a5+7a7+9a9)+(2a2+4a4+6a6+8a 8)]
=(a1+2a2+3a3…+9a9)(a1-2a2+3a3-4a4…+9a9)
=3^10*9
=3^10*3^2
=3^12
两边求导得
9(x+2)^8=a1+2a2x+3a3x^2…+9a9x^8
令x=1,
9(1+2)^8=a1+2a2+3a3…+9a9
9*3^8=a1+2a2+3a3…+9a9
3^2*3^8=a1+2a2+3a3…+9a9
3^10=a1+2a2+3a3…+9a9
a1+2a2+3a3…+9a9=3^10
令x=-1,
9(x+2)^8=a1+2a2x+3a3x^2…+9a9x^8
9(-1+2)^8=a1-2a2+3a3…+9a9
9*1^8=a1-2a2+3a3-4a4…+9a9
9=a1-2a2+3a3-4a4…+9a9
a1-2a2+3a3-4a4…+9a9=9
(a1+3a3+5a5+7a7+9a9)^2-(2a2+4a4+6a6+8a 8)^2
=[(a1+3a3+5a5+7a7+9a9)-(2a2+4a4+6a6+8a 8)][(a1+3a3+5a5+7a7+9a9)+(2a2+4a4+6a6+8a 8)]
=(a1+2a2+3a3…+9a9)(a1-2a2+3a3-4a4…+9a9)
=3^10*9
=3^10*3^2
=3^12
看了已知(x+2)9=a0+a1x...的网友还看了以下:
设随机变量X的分布律为:P(X=k)=,k=1,2,3,…,9,则a=------,P(-1 2020-05-15 …
设x,y为实数且3≤xy^2≤8,4≤x^2/y≤9,则x∧3/y^4的最大值 2020-05-16 …
设x,y为实数,满足3≤xy^2≤8,4≤x^2/y≤9,则x^3/y^4的最大值是 2020-05-16 …
设实数x,y满足3≤xy^2≤8,4≤y^2/x≤9,则x^3/y^4的最大值是 2020-05-16 …
初二数学简单的整式题若2^(2m)=6,2^(2n)=9,则4^(2m-n)=? 2020-05-21 …
填空题有好的答案150分计算:(2^2)(x+3)=计算:(2^2)^2-3^0+(-3)^23x 2020-06-15 …
在等差数列{an}中,a3=2,a7=9,则a15= 2020-07-09 …
一个三角形的两边长分别为2和9,则第三边长为偶数,第三边长为? 2020-07-12 …
设(x+√(x^2+1))(y+√(y^2+4))=9则x√(y^2+4)+y√(x^2+1)=? 2020-10-31 …
若(x+2+m)^9=a0+a1(x+1)^2+.+a9(x+1)^9,且(a0+a2+....+a 2020-10-31 …