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 =

(a) $\frac{1}{n}$

(b) $\frac{n-1}{n}$

(c) $\frac{n+1}{2n}$

(d) $\frac{n+1}{n}$

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,

So, for n terms,

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,

Solving further, we get

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

Therefore,

Hence, the correct option is (d).