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

Question:

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

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

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

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

 

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

Solution:

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,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

So, for terms,

$S_{n}=\frac{n}{2}[2(1)+(n-1) 2]$

$=\frac{n}{2}[2+2 n-2]$

$=\frac{n}{2}(2 n)$

$=n^{2}$.........(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 terms,

$S_{n}=\frac{n}{2}[2(2)+(n-1) 2]$

$=\frac{n}{2}[4+2 n-2]$

 

$=\frac{n}{2}(2+2 n)$

Solving further, we get

$=n(1+n)$

 

$=n^{2}+n$..........(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

$n^{2}+n=k n^{2}$

$k=\frac{n^{2}+n}{n^{2}}$

$k=\frac{n(1+n)}{n^{2}}$

$k=\frac{n+1}{n}$

Therefore, $k=\frac{n+1}{n}$

Hence, the correct option is (d).

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