Here is one way of calculating 1+2+...+n: // Pre-condition: n is 1 or greater // Post-condition: this functions returns the value of // 1+2+3+...+n // Examples: sum(1) is 1, sum(2) is 3, sum(3) is 6. ... int sum (int n) int sum = 0; while (n > 0) { sum sum + n; --n; } return sum; } Each question below will use the same header, so all you need to write is the code that would appear in the function body. Do not change the function header, or the behavior of the function. a) Rewrite the body of the sum function using a single for-loop instead of a while-loop. b) Rewrite the body of the sum function using a single do-while loop instead of a while-loop. (4 marks) c) Rewrite the body of the sum function using recursion instead of a while-loop. (4 marks) d) Rewrite the body of the sum function using no loops or recursion. [hints: 1+2+3+4+.....+n= n(n+1)/2] (6 marks)