Sum of the First n Natural Numbers We prove the formula 1+ 2+ + n = n(n+1) / 2, for n a natural number.

Step by step descriptive logic to print even numbers from 1 to n without using if statement.

1. Input upper limit to print even number from user.
2. Run a loop from first even number i.e. 2 (in this case), that goes till n and increment the loop counter by 2 in each iteration.
3. Finally inside loop body print the value of i .

The even numbers from 1 to 100 are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98 and 100.

Sum of first n even numbers = n * (n + 1).

Sum of First Ten Even numbers

For n Natural Numbers The family of natural numbers includes all the counting numbers, starting from 1 till infinity. If n consecutive natural numbers are 1, 2, 3, 4, …, n, then the sum of squared ‘n’ consecutive natural numbers is represented by 12 + 22 + 32 + … + n2.

Hence this series is an Arithmetic Progression. First term of this sequence is 1 and the common difference is 2. So we get our sequence is arithmetic progression with a = 1 and d = 2. Hence the sum of the first n odd natural number is n2.

Example: Prove that the sum of the n first odd positive integers is n2, i.e., 1 + 3 + 5 + ··· + (2n − 1) = n2.

Explanation: The primes, in order, are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53.