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

It forms an AP where

first term a = 2

common difference d = 4-2 = 2

Now sum of n terms  = (n/2)*{2a + (n-1)d}

                                 = (n/2)*{2*2 + (n-1)2}

                                 = n*{2 + (n-1)} 

                                 = n(n + 1) 

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

It forms an AP where

first term a = 1

common difference d = 3-1 = 2

Now sum of n terms  = (n/2)*{2a + (n-1)d}

                                 = (n/2)*{2*1 + (n-1)2}

                                 = n*{1+ (n-1)} 

                                 = n(n + 1 - 1)

                                 = n*n

                                 = n²

Now according to question,

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

=> n(n+1) = k*n²

=> k*n = n+1

=> k = (n+1)/n

So value of k is (n+1)/n