Show that the sum of first n even natural numbers is equal to 1 + 1 n times the sum of first n odd natural numbers.

Given: Even natural numbers A.P. 2 , 4 , 6 , 8 , . . . . . . Odd natural numbers A.P. 1 , 3 , 5 , 7 , . . . . . . We know that S n = n 2 2 a + n - 1 d Finding the sum of first n even natural numbers we have a = 2 , d = 4 - 2 = 2 ⇒ S n = n 2 2 × 2 + n - 1 2 ⇒ S n = n 2 4 + 2 n - 2 ⇒ S n = n 2 2 + 2 n ⇒ S n = n 2 × 2 1 + n ⇒ S n = n + n 2 . . [ 1 ] Sum of first n odd natural numbers we have a = 1 , d = 3 - 1 = 2 ⇒ S n = n 2 2 × 1 + n - 1 2 ⇒ S n = n 2 2 + 2 n - 2 ⇒ S n = n 2 × 2 n ⇒ S n = n 2 . . [ 2 ] Sum of first n even natural numbers = 1 + 1 n times the sum of first n odd natural numbers. From equation 1 and 2 ⇒ n + n 2 = 1 + 1 n × n 2 ⇒ n + n 2 = n 2 + n 2 n ⇒ n + n 2 = n + n 2 L . H . S = R . H . S Hence proved.