Program #2 Define a sequence of numbers recursively Define a sequence as, at, az, as, where ao = 1 an = (an-1+ 1) 2 an = (2* an - 2 + aa -1) ifn is even So for instance: (n = 0) ao = 1 (n = 1) a; = (ao + 1) * 2 = (1+ 1)*2= 4 (n = 2) az = (2* ao + ao) = 2*1+4 = 6 (n = 3) as = (az + 1) *2= (6+ 1) * 2 =14 The resulting sequence is 1, 4, 6, 14, - %3D if n is odd %3D You will write a function that returis the nth term of the sequence. You must do this by using a recursive function. The function will NOT print out any values. Rather, the function will return the nth term of the sequence using a recursive algorithm. In the main part of the program: 1. Input the number of terms of the sequence to output. 2. In a for loop, call the function repeatedly to get the desired number of terms. The function will take i (assuming the for index is called i) as the argument and return the ith term of the sequence. 3. In the for loop, print out each term as it is returned. Figure out the necessary prompts for the inputs and the desired outputs by looking at this example session. The number in red is a possible input and is not what you print out. Enter the number of terma> 4 Term 10> 1 Term 1> 4 Term 12> 6 Term 13> 14 HINTS: 1. The base case is when n==0 2. In the recursive case you will need to decide if n is odd or even. n is odd if there is a remainder when you divide by two. if (n % 2) !=0): #test for odd Since a nonzero number is true, the above could be shortened to: if (n % 2): #test for odd Either way, else: #must be even

Write a C++ program using classes and recursion functions to convert a number in a given base (the user enters the base - from 2 to 9) to a decimal number. The algorithm states that each successive digit in the number is multiplied by the base raised to the power corresponding to its position in the number. The low-order digit is in position 0. We sum together all of these products to get the decimal value. For example, in binary – base 2, if we have the binary number 111001, we convert it to decimal as follows: 1x 25+1x 24+1x23+0x2²+0x 21+1x 2º=57 For a base 5 number, 14230: 1x54+4x53+2x52+3x51+0 x 50=1190 Validate your input. You should check for invalid digits and provide an informative error message to the user. This program requires you to use a recursive function for the conversion to decimal. Failure to do that will result in no points awarded.

Write a C++ program using classes and recursion functions to covert a number in a given base (the user enters the base - from 2 to 9) to a decimal number. The algorithm states that each successive digit in the number is multiplied by the base raised to the power corresponding to its position in the number. The low-order digit is in position 0. We sum together all of these products to get the decimal value. For example, in binary- base 2, if we have the binary number 111001, we convert it to decimal as follows: 1x25 + 1x2* + 1x2³ + 0x2²+ 0x2² + 1x2º = 57 For a base 5 number, 14230: 1x5 + 4x53 + 2x52 + 3x5+ Ox5° = 1190 Validate your input. You should check for invalid digits and provide an informative error message to the user. This program requires you to use a recursive function for the conversion to decimal. Failure to do that will result in no points awarded.

Consider a recursive function, called f, that computes powers of 3 using only the + operator. Assume n > = 0. int f(int n) { if (n == 0) return 1; return f(n-1) + f(n-1) + f(n-1); } Give an optimized version of f, called g, where we save the result of the recursive call to a temporary variable t, then return t+t+t. i got int g(int n) { if (n == 0) return 1; int t = g(n - 1); return t+t+t; } so now Write a recurrence relation for T(n), the number addition operations performed by g(n) in terms of n.