-n + 4(n + 1) = 3n+4
n^2 + n
If you mean: -6n+5 = 1 then the value of n is 2/3
It means that a + n + mc + cg = 0
1. With boolean algebra, 1 + n is always equal to 1, no matter what the value of n is.
2 plus n is the true one, but 1 plus n is not?
Did you mean 1-2+3-4+5-6....till n ?? If yes, then here it is: #include<iostream.h> void main() { int n,sum=0,check=1; cout<<"Enter n "; cin>>n; for(int i=1;i<=n;i++) { sum+=(check*i); check*=-1; } cout<<"The result of series is: "<<sum; }
-n + 4(n + 1) = 3n+4
They have n in common.
if (n%2==0) sum=n/2*(n+1); else sum=(n+1)/2*n;
(n2 + n)/2 1+2+3+4+5+n= 15+n
n^2 + n
If you mean: -6n+5 = 1 then the value of n is 2/3
Successor of n=n+1 Predecessor of n=n-1 Sum=[n+1]+[n-1] Plus 1 plus minus 1= [1]+[-1]=0 [n][n] =2n Hope this helps
n = 28
int n, i; for(n = 1; n <= 5; ++n) { for(i = 1; i <= n; ++i) { printf("%d", n); } }
It means that a + n + mc + cg = 0