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);
    }

a. geometricb. g[h] = (3/4)·g[h-1]; g[1] = 384c. g[h] = 384·(3/4)^(h-1)d. g[12] ≈ 16.2183 gallonse. g[12] ≈ 64.8732 quarts

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt