Question

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

Options

• \[\frac{1}{n}\]
   
• \[\frac{n - 1}{n}\]
   
• \[\frac{n + 1}{2n}\]
   
• \[\frac{n + 1}{n}\]

   

Answer 1

In the given problem, we are given that the sum of the first n even natural numbers is equal to k times the sum of first n odd natural numbers.

We need to find the value of k

Now, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 1

Common difference (d) = 2

So, let us take the number of terms as n

Now, as we know,

`S_n = n/2 [2a + (n-1) d]`

So, for n terms,

`S_n = n/2 [ 2(1) + (n-1)2]`

    `= n/2 [ 2 + 2n - 2]`

   `=n/2 (2n)`

    = n²                        ..................(1)

Also, we know that the first even natural number is 2. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 2

Common difference (d) = 2

So, let us take the number of terms as n

So, for n terms,

`S_n = n/2 [ 2 (2) + (n - 1) 2]`

     `= n/2 [4 + 2n - 2 ]`

     ` = n /2 ( 2 + 2n)`

Solving further, we get

= n ( 1 + n) 

= n² + n                               .................(2) 

Now, as the sum of the first n even natural numbers is equal to k times the sum of first n odd natural numbers

Using (1) and (2), we get

 n² + n = kn²   

         ` k =(n^2 + n )/(n^2)`

         `k = (n(1 +n))/(n^2)`

         `k = ( n+ 1)/n`

Therefore,   `k = ( n+ 1)/n`