Question 5  Q9A. Consider the following algorithm: sum = 0 for j in range(1,15):     sum = sum + (6*j - 4)      print(sum) What is printed as a result of executing this algorithm?

Question 1 Q9. Consider the following algorithm: sum = 0 for j in range(1,14):    sum = sum + (4*j + 6)    print(sum) What is printed as a result of executing this algorithm?

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sum = 0 for j in range(1,12):    sum = sum + (9*j + 3)    print(sum) What is printed as a result of executing this algorithm?

n-1 Geometric (n) = i=1 i=1 1 1 * n-1 П %3D Harmonic (n) = i=1 n Let's look at examples. If we use n = 4, the geometric progression would be 1 * 2 * 3 * 4 = 24, and the harmonic 1.1.1 1 progression would be 1* -= 0.04166. 2 3 4 24 Task #1 Tracing Recursive Methods 1. Copy the file Recursion.java (see Code Listing 16.1) from the Student Files or as directed by your instructor. 2. Run the program to confirm that the generated answer is correct. Modify the factorial method in the following ways: a. Add these lines above the first if statement: int temp; System.out.println ("Method call -- " + "calculating " "Factorial of: " + n); + Copyright © 2019 Pearson Education, Inc., Hoboken NJ b. Remove this line in the recursive section at the end of the method: return (factorial (n - 1) * n); c. Add these lines in the recursive section: temp = factorial (n - 1); System.out.println ("Factorial of: " (n - 1) + " is " temp); return (temp * n); 3. Rerun the program and note how the recursive calls…

The following code is used to derive Fibonacci algorithm with this sequence: (0, 1, 1, 2, 3, 5, 8, 13, 21, …) where each number is “add” of two previous ones. int fib(int n){ if (n==0) return 0; else if (n == 1) return 1; else return fib(n−1) + fib(n−2); } Explain why the following RISC-V assembly code works, you need to explain all the details using the comments as hints?   # IMPORTANT! Stack pointer must remain a multiple of 16!!!!   addi x10,x10,8                # fib(8), you can change it fib: beq x10, x0, done             # If n==0, return 0 addi x5, x0, 1 beq x10, x5, done             # If n==1, return 1 addi x2, x2, -16              # Allocate 2 words of stack space sd x1, 0(x2)                  # Save the return address sd x10, 8(x2)                 # Save the current n addi x10, x10, -1             # x10 = n-1 jal x1, fib # fib(n-1) ld x5, 8(x2) # Load old n from the stack sd x10, 8(x2) # Push fib(n-1) onto the stack addi x10, x5, -2 # x10 = n-2 jal x1, fib # Call fib(n-2)…