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. 1/n
B. n-1/n
C. n+1/2n
D.n+1/n

Answer 1

Answer:

The value of k is $\dfrac{n-1}{n}$  

Step-by-step explanation:

Given as :

Statement

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

According to question

Let The n even natural numbers = 2 , 4 , 6 ,   ......... 2 n

So, The sum of n even natural numbers = $S_n_1$ = $\dfrac{n}{2}$  [ 2 a + ( n - 1 ) d ]

where a is first terms , d is common difference

So, a = 2

     d = 2

∴   $S_n_1$ = $\dfrac{n}{2}$  [ 2 × 2 + (  n - 1 ) × 2 ]

Or, $S_n_1$ = $\dfrac{n}{2}$  [ 4 + 2 n - 2 ]

or, $S_n_1$ = $\dfrac{n}{2}$ ( 2 + 2 n)

i.e  $S_n_1$ =  n ( n + 1 )

So, The sum of n even natural numbers = $S_n_1$ = n ( n + 1 )

Again

Let The n odd natural numbers = 1 , 3 , 5 ,   .........  2 n - 1

So, The sum of n odd natural numbers = $S_n_2$ = $\dfrac{n}{2}$  [ 2 a + ( n - 1 ) d ]

where a is first terms , d is common difference

So, a = 1

     d = 2

∴   $S_n_2$ = $\dfrac{n}{2}$  [ 2 × 1 + ( n - 1 ) × 2 ]

Or, $S_n_2$ = $\dfrac{n}{2}$  [ 2 + 2 n - 2 ]

Or, $S_n_2$ = $\dfrac{n}{2}$ [ 2 n ]

Or, $S_n_2$ =  n²

i.e  $S_n_2$ =  n²

So, The sum of n odd natural numbers =  $S_n_2$ =  n²

Now, According to question

The sum of first n even natural numbers =  k × The sum of first n odd natural numbers

Or, $S_n_1$  =  k $S_n_2$

i.e   n ( n + 1 ) = k × n²

Or, k = $\dfrac{n(n-1)}{n^{2} }$

∴    k = $\dfrac{n-1}{n}$

So, The value of k = $\dfrac{n-1}{n}$

Hence, The value of k is $\dfrac{n-1}{n}$   Answer