计算：i 2n-3+ i 2n-1+ i 2n+1 +i 2n+3 (n大于等于2,且n属于N)其中 2n-3 ,2n-1,2n+1,2n+3 是i的幂

计算:1+i+i^2+i^3+...+i^n 计算：1+i^n+i^2n+......+i^2000n(n属于N+) 1+i^n+i^2n+...+i^2000n＝? 计算(1+i)+(2+3i)+(4+5i)+…+[(2∧n-1)+(2n-1)i] n属于r 求1+i^n+i^2n+i^3n 高二文科复数运算!i^4n+i^4n+1+i^4n+2+i^4n+3 计算：i 2n-3+ i 2n-1+ i 2n+1 +i 2n+3 (n大于等于2,且n属于N)其中 2n-3 ,2n-1,2n+1,2n+3 是i的幂 i^4n=1,i^(4n+1)=i,i^(4n+2)=-1,i^(4n+3)=-i,当n∈Z时,能成立吗,书上是n∈N 复数 设a=(1+√3i)/2,b=(1-√3i)/2,当n∈N*时,计算a^n+b^n (1+i)^n+(1+i)^n-1+(1+i)^n-2+……(1+i)求计算过程及结果公式, 计算(i^2+i)+|i|+(1+i) 计算下列值 急……(1-i)^2n/(1+i)^2m（m,n∈N） (1+i)^2n/1-i+(1-i)^2n/1+i=2^n,求最小正整数n? 求n项和数列极限,通式为i/(i+n)就是n->无穷,1/(n+1)+2/(n+2)+3/(n+3)+...+n/2n; 编程计算：n ∑(i+1)/i!i=1（精度要求为(n+1)/n! #define N 20 fun(int a[],int n,int m) {int i; for(i=m;i>n;i--)a[i+1]=a[i]; return m; } void main() #define N 20fun(int a[],int n,int m){int i;for(i=m;i>n;i--)a[i+1]=a[i];return m;}void main(){ int i,a[N]={1,2,3,4,5,6,7,8,9,10};fun(a,0,N/2);for(i=0;i 计算e^i/n+e^2/n+...+e^n/n=? 关于lingo程序的,sets:A/1..4/:c,b,n;X/A,A/:D;endsetsmin=1.1*n(1)+1.2*n(2)+1.3*n(3)+1.4*n(4);@for(A(i):@sum(A(j):c(i)*D(i,j)=1750);@for(A(i):@sum(A(j):n(i)*D(i,j)=b(i));@for(A(i):@gin(n(i)));@for(A(j):@for(A(i):@gin(D(i,j))));n(1)>n(2);n(2)>n(3);n