Question

If the sum of first n even natural numbers is equal to k times

Answer 1

Step-by-step explanation:

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

It forms an AP where

first term a = 2

here we can see that the common difference between numbers = 2

let's do

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 = 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

= n^2

According to the problem :

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

=> n(n+1) = k*n^2

=> k*n = n+1

=> k = (n+1)/n