Question

Write a function that is passed in a positive integer, n, and sum up all the digits from 1 to n, doubling the even numbers. In other words, calculate and return the following sum. if n is odd: 1 (2 * 2) 3 (4 * 2) ... ((n - 1) * 2) n if n is even: 1 (2 * 2) 3 (4 * 2) ... (n - 1) (n* 2)

Answer 1

Answer:

Written in Python

def sums(n):

    total = 0

    for i in range(1,n+1):

         if i%2 == 0:

              total = total + i*2

         else:

              total = total + i

    return(total)

Step-by-step explanation:

The function is written in Python

def sums(n):

This initializes total to 0

    total = 0

This iterates through the integers that makes up n i.e. 1,2,3.....n

    for i in range(1,n+1):

This checks if current digit is even

         if i%2 == 0:

If yes, it adds the double of the digit

              total = total + i*2

         else:

If otherwise, it adds the digit

              total = total + i

This prints the calculated sum or total

    return(total)

To call the function from main, use the following:

print(sums(n))

Where n is a positive integer. i.e. $n \geq 1$