Write a program using recursion that computes the sum of the series: 1^2+5^2+ 6^2 + 11^2 + 17^2+28^2 +... + n^2 (n is an input)
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Q: Write a program using recursion that computes the sum of the series: 1^2 + 5^2 + 6^2 + 11^2 + 17^2 +…
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- Write a program using recursion that computes the sum of the series: 1^2 + 5^2 + 6^2 + 11^2 + 17^2 + 28^2 +… …. + n^2 (n is an input)Use the recursion to solve the following problems using Java Programming Language 1. Calculate the sum of an array of n integers. 2. Compute Powers, p(x,n)=xn 3. Revers an array of n integers 4. Calculate the length of string (number of characters) 5. Implement the tail recursion for problems 1 and 2. Please test the program by using 2 different input sets!Ackermann’s function is a recursive mathematical algorithm that can be used to test how well a computer performs recursion. Write a function A(m, n) that solves Ackermann’s function. Use the following logic in your function: If m = 0 then return n + 1 If n = 0 then return A(m-1, 1) Otherwise, return A(m-1, A(m, n-1)) Test your function in a driver program that displays the following values:A(0, 0) A(0, 1) A(1, 1) A(1, 2) A(1, 3) A(2, 2) A(3, 2)
- Write a recursive MIPS program that computes this sequence of numbers (0, 1,3,9, 35, 169,987, 6769, 53307, 473841, 4691027, ....) can be computed with the following recursive function: int funt (int n) { if (n == 0) return 0; else if (n == 1) return 1; else return (n-1) * funt (n-1) + (n-2) *funt (n-2) + 2; } Thus, funt(0) = 0, funt(1) = 1, funt(2) = 3, funt(3) = 9, funt(4) = 35, aren't off by 1 in the sequence! Your main code is main () { } cout> n; cout<<"The number series of n numbers is: "; for (int i = 0; i <= n; i++) { cout<Write a recursive function to compute the following series: m(i) = 1/3 + 2/ 5 + 3/7 + 4/ 9 + 5/11 + 6/13 + . . . + i/ 2i + 1 Write a test program that displays m(i) for i 1, 2, ..., 10.java C++ Ackermann’s FunctionAckermann’s Function is a recursive mathematical algorithm that can be used to test how well a computer performs recursion. Write a function A(m, n) that solves Ackermann’s Function. Use the following logic in your function:If m = 0 then return n + 1If n = 0 then return A(m−1, 1) Otherwise, return A(m−1, A(m, n−1))Test your function in a driver program that displays the following values:A(0, 0) A(0, 1) A(1, 1) A(1, 2) A(1, 3) A(2, 2) A(3, 2) SAMPLE RUN #0: ./AckermannRF Hide Invisibles Highlight: Show Highlighted Only The·value·of·A(0,·0)=·1↵ The·value·of·A(0,·1)=·2↵ The·value·of·A(1,·1)=·3↵ The·value·of·A(1,·2)=·4↵ The·value·of·A(1,·3)=·5↵ The·value·of·A(2,·2)=·7↵ The·value·of·A(3,·2)=·29↵function recursion(B[0..n − 1], i) if n == 0 then return False if n == 1 then return (B[0] == i)x ← recursion(B[0........n/2 − 1], i) y ← recursion(B[n/2..........4 × n/2 − 1], i) z ← recursion(B[4 × n/2..........n − 1], i) return (x OR y OR z) input array is of length n = 2^p, p is a positive integer Write the recursive formula for above algorithm as of worst case inputs.Using recursion, write a Java program that takes an input ‘n’ (a number) ( user should gives the n value if user asks fibonacci series of 20 then it should display 20 numbers of fibonacci)from a user to calculate and print out the Fibonacci using the following modified definition: F(N) = 1 if n = 1 or n = 2 = F((n+1)/2)2 + F((n-1/2)2 if n is odd = F(n/2 + 1)2 – F(n/2 – 1)2 if n is evenRECURSIVE PYTHON The Fibonacci sequence begins with 0 and then 1 follows. All subsequent values are the sum of the previous two, for example: 0, 1, 1, 2, 3, 5, 8, 13. Complete the fibonacci() function, which takes in an index, n, and returns the nth value in the sequence. Any negative index values should return -1. Ex: If the input is: 7 the output is: fibonacci(7) is 13 Note: Use recursion and DO NOT use any loops. # TODO: Write recursive fibonacci() functiondef fibonacci(): if __name__ == "__main__": start_num = int(input()) print('fibonacci({}) is {}'.format(start_num, fibonacci(start_num)))Compute f(6) for the recursive function below. def f(n): if n == 0: return 1 if n == 1: return 2 else: return f(n-1)+n*f(n-2)-nImplement a recursive program that takes in a number and finds the square of that number through addition. For example if the number 3 is entered, you would add 3+3+3=9. If 4 is entered you would add 4+4+4+4=16. This program must be implemented using recursion to add the numbers together. I need the MIPS Code for the above.Using recursion, write a Java program that takes an input ‘n’ (a number) from a user to calculate and print out the Fibonacci using the following modified definition: F(N) = 1 if n = 1 or n = 2 = F((n+1)/2)2 + F((n-1/2)2 if n is odd = F(n/2 + 1)2 – F(n/2 – 1)2 if n is even Your solution must implement recursion to receive points for this question.SEE MORE QUESTIONS