When any number which ends with 0,2,4,6,8 is divided by 2 that is an even number. And when any number ends with 1,3,5,7,9 is not divided by two is an odd number.
Example:
Input : 8
Output: Sum of First 8 Even numbers = 72
Sum of First 8 Odd numbers = 64
Approach #1: Iterative
- Create two variables evenSum and oddSum and initialize them by 0.
- Start For loop from 1 to 2*n.
- If i is even Add i with evenSum.
- Else add i with oddSum.
- Print evenSum and oddSum at the end of loop.
Below is the implementation of the Java program:
Java
// Calculate the Sum of First N Odd & Even Numbers in Javaimport java.io.*;public class GFG { // Driver function public static void main(String[] args) { int n = 8; int evenSum = 0; int oddSum = 0; for (int i = 1; i <= 2 * n; i++) { // check even & odd using Bitwise AND operator if ((i & 1) == 0) evenSum += i; else oddSum += i; } // Sum of even numbers less than 17 System.out.println("Sum of First " + n + " Even numbers = " + evenSum); // sum of odd numbers less than 17 System.out.println("Sum of First " + n + " Odd numbers = " + oddSum); }} |
Output
Sum of First 8 Even numbers = 72 Sum of First 8 Odd numbers = 64
Time Complexity: O(N), where N is the number of First N even/odd numbers.
Auxiliary Space: O(1)
Method 2: Using AP Formulas.
- Sum of First N Even Numbers = n * (n+1)
- Sum of First N Odd Numbers = n * n
Below is the implementation of the above approach:
Java
// Calculate the Sum of First N Odd & Even Numbers in Javaimport java.io.*;public class GFG { // Function to find the sum of even numbers static int sumOfEvenNums(int n) { return n * (n + 1); } // Function to find the sum of odd numbers. static int sumOfOddNums(int n) { return n * n; } // Driver function public static void main(String[] args) { int n = 10; int evenSum = sumOfEvenNums(n); int oddSum = sumOfOddNums(n); // Sum of even numbers System.out.println("Sum of First " + n + " Even numbers = " + evenSum); // sum of odd numbers System.out.println("Sum of First " + n + " Odd numbers = " + oddSum); }} |
Output
Sum of First 10 Even numbers = 110 Sum of First 10 Odd numbers = 100
Time Complexity: O(1)
Auxiliary Space: O(1)
