Can someone explain how this recursive function output is 12? Recursion is hard for me

Alert dont submit AI generated answer. Explain in details, please

Given the recursive function definition:t(n) = n+ 3 * t(n-1) + t(n-2) t(0) = 2 t(1) = 1a. Evaluate: t(3)b. Write the source code to implement this function

Can someone explain the trace of this resursion problem? It is confusing to me def r(n):    if n>=5:        return 10    return r(n+1)+2print(r(1))

Write a recursive method that returns the value of N! (N factorial) using the definition given in this chapter. Explain why you would not normally use recursion to solve this problem.

Make a code using Recursion The countSubstring function will take two strings as parameters and will return an integer that is the count of how many times the substring (the second parameter) appears in the first string without overlapping with itself. This method will be case insensitive.   For example: countSubstring(“catwoman loves cats”, “cat”) would return 2 countSubstring(“aaa nice”, “aa”) would return 1 because “aa” only appears once without overlapping itself.     public static int countSubstring(String s, String x) { if (s.length() == 0 || x.length() == 0) return 1;   if (s.length() == 1 || x.length() == 1){ if (s.substring(0,1).equals(x.substring(0,1))){ s.replaceFirst((x), " "); return 1 + countSubstring(s.substring(1), x); } else { return 0 + countSubstring(s.substring(1), x); } } return countSubstring(s.substring(0,1), x) + countSubstring(s.substring(1), x); }   public class Main { public static void main(String[] args) {…

write a code in java (using recursion)

Write a recursive function to determine if an array of integers contains any even numbers: bool hasEvens(int nums[], int size)

Write a recursive function (Java) called Fac which takes one positive integer argument (n) and returns n! You may not use a built-in factorial method or function.

Which is the base case of the following recursion function: def mult3(n):    if n == 1:        return 3    else:        return mult3(n-1) + 3     else     n == 1     mult3(n)     return mult3(n-1) + 3