C(N,1)COSX+C(N,2)COS2X+-----+C(N,N)COSNX
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/27 21:58:28
C(N,1)COSX+C(N,2)COS2X+-----+C(N,N)COSNXC(N,1)COSX+C(N,2)COS2X+-----+C(N,N)COSNXC(N,1)COSX+C(N,2)COS
C(N,1)COSX+C(N,2)COS2X+-----+C(N,N)COSNX
C(N,1)COSX+C(N,2)COS2X+-----+C(N,N)COSNX
C(N,1)COSX+C(N,2)COS2X+-----+C(N,N)COSNX
设 C(N,1)COSX+C(N,2)COS2X+-----+C(N,N)COSNX=A ,
C(N,1)sinX+C(N,2)sin2X+-----+C(N,N)sinNX=B
则 A+Bi= C(N,1)e^iX+C(N,2)e^2iX+-----+C(N,N)e^Nix
=(1+cosx+isinx)^n-1=[2cos^20.5x+2isin0.5xcos0.5x]^n-1
=2^ncos^n0.5x[cos0.5nx+isin0.5nx]-1
所以 A=2^ncos^n0.5xcos0.5nx-1,B=2^ncos^n0.5xsin0.5nx
C(N,1)COSX+C(N,2)COS2X+-----+C(N,N)COSNX
∫[1/cos^2(x)]+1 d(cosx) 等于 A(-1/cosx)+cosx+C B (1/cosx)+cosx+C C(-cotx)+cosx+C D cotx+cosx+C
求lim(x→0) (cosx+cos^2+.+cos^n-n)/cosx-1 不能用洛必达法则求lim(x→0) (cosx+cos^2+.+cos^n-n)/cosx-1 不能用洛必达法则
证明:c(n,0)c(n,1)+c(n,1)c(n,2)+...c(n,n-1)c(n,n)=c(2n,n-1)
(-1)^n*cosx=cos(x+nπ)
C(n,0)+C(n,1)+C(n,2)+…+C(n,n-2)+C(n,n-1)+C(n,n)为什么等于什么
matlab解方程组方程组1:(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2-n*sin(c/2+d)+e*cos(f-b))^2-(h+n*cos(c/2+d)-e*sin(f-b))^2=0;方程组2:(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2+e*cos(a+f)-n*sin(c/2-d))^2-(h
2cosxcos(n-1)x等于什么?cosx+cos(n-2)x
求和C(n,1)+2^2C(n,2)+.+n^2C(n,n)=?
组合:C(n,0)+C(n,1)+……+C(n,n)=n^2
C(0,n)+2C(1,n)+3C(2,n)+...+(r+1)C(r,n)+...+(n+1)C(n,n)=___(n属于N*)
求证:C(0,n)+2C(1,n)+.+(n+1)C(n,n)=2^n+2^(n-1)
证明C(0,n)^2+C(1,n)^2+……+C(n,n)^2=C(n,2n)
求Sn=C(n,1)+2C(n,2)+...+nC(n,n)C(n,1)+2C(n,2)+...+nC(n,n) n是下标
如何证明C(0,n)+C(1,n)+C(2,n)+.+C(n-1,n)+C(n,n)=2的N次方 不用数学归纳法
lim (n趋近正无穷) cos x /2cosx/4…cosx/2^n
C(n.0)+2C(n.1)+4C(n.2)+C(n.2)+C(n.3)…+C(n.n)=?
求证c(n,1)+2c(n,2)+3c(n,3)+...+nc(n,n)=n2^(n-1)