作业帮A(1+β)n –X[1+(1+β)+(1+β)2+…+(1+β)n-1]= A(1+β)n –X[(1+β)n-1]/β 是怎么推导出来的啊
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/30 09:17:12
![A(1+β)n –X[1+(1+β)+(1+β)2+…+(1+β)n-1]= A(1+β)n –X[(1+β)n-1]/β 是怎么推导出来的啊](/uploads/image/z/11663467-43-7.jpg?t=A%281%2B%CE%B2%29n+%E2%80%93X%5B1%2B%281%2B%CE%B2%29%2B%281%2B%CE%B2%292%2B%E2%80%A6%2B%281%2B%CE%B2%29n-1%5D%3D+A%281%2B%CE%B2%29n+%E2%80%93X%5B%281%2B%CE%B2%29n-1%5D%2F%CE%B2+%E6%98%AF%E6%80%8E%E4%B9%88%E6%8E%A8%E5%AF%BC%E5%87%BA%E6%9D%A5%E7%9A%84%E5%95%8A)
x)s0>I3OQhCmJi?jXe**Gۤlg
}Ovv<[t]|Vө]6IEJΆzڻŲmC*Nh&6qP xڱv>[?
O;l{pI=O[
A(1+β)n –X[1+(1+β)+(1+β)2+…+(1+β)n-1]= A(1+β)n –X[(1+β)n-1]/β 是怎么推导出来的啊
A(1+β)n –X[1+(1+β)+(1+β)2+…+(1+β)n-1]= A(1+β)n –X[(1+β)n-1]/β 是怎么推导出来的啊
A(1+β)n –X[1+(1+β)+(1+β)2+…+(1+β)n-1]= A(1+β)n –X[(1+β)n-1]/β 是怎么推导出来的啊
即是要证[1+(1+β)+(1+β)^2+…+(1+β)^(n-1)]= [(1+β)^n-1]/β
是利用等比数列的前n项和公式求得的
[1+(1+β)+(1+β)^22+…+(1+β)^(n-1)]= 1[1-(1+β)^n]/[1-(1+β)]=[(1+β)^n-1]/β
A(n,n)+A(n-1,n-1)=XA(n+1,n+1)求X
A(1+β)n –X[1+(1+β)+(1+β)2+…+(1+β)n-1]= A(1+β)n –X[(1+β)n-1]/β 是怎么推导出来的啊
[x^n(a-1)^2n]/[x^2n(a-1)^n+1]分式化简,
证明：x^n-na^(n-1)x+(n-1)a^n能被(x-a)^2整除(n>=2,n属于N*)
（a+x)^(m+1)*(b+x)^(n-1)-(a+x)^m*(b+x)^n
3ax^n+1-6a^2x^n+3a^2x^n-1因式分解
若集合A={x|x=3n+1,n∈N},B={x|x=3n-2n∈N},则A∩B
因式分解：9a^2+x^2n+6a+2x^n+6ax^n+1
9a^2+x^2n+6a+2x^n+6ax^n+1 因式分解
化简a^2n-6`(a^2)^3+(x-a^n)(x+a^n)-(x-1)^2
因式分解；a^4x^n+2-4x^n （n为正整数）=多少因式分解；25a^n+2-10a^n+1+a^n=多少?
f(x)=e^x-x 求证(1/n)^n+(2/n)^n+...+(n/n)^n
(x²-x+1)^n=a0+a1x+a2x²+...+a(2n)x^(2n) n∈N*,则a1+a2+a3+...+a(2n-1)=
矩阵A m×n,矩阵 X n×1,m
集合A={x|x=2n+1,n属于N},B={x|x=4n+1,n属于N}之间的关系
n为正整数,计算a^5n*（-a）^3n-（-a）^4n*a^4n若x*x^m*x^n=x^12,且m比n大1,求m,n的值
lim(x->0)[cosx-1+1/2 *(x^2)]/x^n=a 求 n ,a
x^(n)*x^(n+1)+x^(2n)*x