Write a for loop that assigns summedvalue with the sum of all odd values from 1 to usernum. assume usernum is always greater than or equal to 1.

// Input value is usernum.
// This code snippet sums 1 + 3 + 5 + ... + usernum
//  The answer is stored in the variable summedvalue.

N = (int) (usernum+1)/2;    // maximum number of integers to be summed
int *v = malloc(N*sizeof(int));     //  allocate storage for array v

// Calculate the number of loop counts and assign array v..
count = 0;
k = 1;
while (1) {
       if (k>usernum) {   //  do not extend v beyond usernum
          break;
          }
       v(count) = k;        //  assign an odd integer to v, including usenum
       count++;
       k += 2;                //  k is an odd number
       if k>usernum {          // handle usernum as odd or even
          k = usernum;
          } 
 }
n = count;          // the size of array v.

// Calculate the sum in a for loop
summedvalue  = 0;               // initialize summedvalue
for (i=0; i<=n; i++) {
     summedvalue += v(i);
    }

Hey mate here is the answerthe correct answer is b may it will you