The Fibonacci series can be computed as follows, F(n) = F(n 1) + F(n 2) (1) In class, we showed how this can be done in O(log n) computation time. Now suppose that the de nition is changed in the following way, F0(n) = F0(n 1) + F0(n 2) + F0(n 3) + F0(n 4) (2) Can F0(n) be computed in O(log n)? If yes, please show how it can be done. If no, show a counterexample where this fails. Please provide your rationale for both. Assume that F0(0) = 0; F0(1) = 1; F0(2) = 1; F0(3) = 1.

Aprovides legal protection for something an individual has created

Chloe is building a kiosk-based excel application. she wants to make some modifications to the screen elements in order to keep users from being distracted by parts of the application that are irrelevant to her application. she turns to henry for guidance as she knows he built a similar solution earlier this year.chloe has decided to hide the worksheet gridlines and the vertical scroll bar. what does henry tell her to use to do this? a) screen elements dialog boxb) display options dialog boxc) customization dialog boxd) excel options dialog box

Write a pseudocode solution for each of these problems. 1. design a while loop that lets that user enter a number. the number should be multiplied by 10, and the result stored in a variable named product. the loop should iterate as long as product contains a value less than 100. 2. design a do-while loop that asks the user to enter two numbers. the numbers should be added and the sum displayed. the loop should ask the user whether he or she wishes to perform the operation again. if so, the loop should repeat; otherwise it should terminate. 3. design a for loop that displays the following set of numbers: 0, 10, 20, 30, 40, 50 100. 4. design a nested loop that displays 10 rows of # characters. there should be 15 # characters in each row. 5. convert this for loop to a while loop. declare integer count for count = 1 to 50 display count end for 6. find the error in the following pseudocode. declare boolean finished = false declare integer value, cube while not finished display “enter a value to be cubed.” input value; set cube = value ^ 3 display value, “ cubed is “, cube end while