If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k =

First n even natural numbers are: 2, 4, 6, 8, ............

It forms an AP where first term a = 2 and common difference d = 4 – 2 = 2

Now sum of n terms = () X {2a + (n–1)d}

= (() x {2 x 2 + (n–1)2}

= n x {2 + (n–1)}

= n (n + 1)

First n odd natural numbers are: 1, 3, 5, 7, ............

It forms an AP where first term is a = 1 and the common difference d = 3 – 1 = 2

Now sum of n terms = () x {2a + (n–1) d}

= () x {2 x 1 + (n–1)2}

= n x {1 + (n–1)}

= n (n + 1 – 1)

= n x n

= n²

Now, According to given condition

Sum of first n even numbers = k X (Sum of first n odd numbers)

n (n + 1) = k x n²

k x n = n + 1

k =