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

Answer:

The value of k is (n + 1)/n .

Among the given options option (d) (n + 1)/n is a correct answer.

Step-by-step explanation:

Given:  

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

 

First n even natural numbers are 2,4 ,6 ,………., 2n  which forms an AP.

Here, a = 2, d = 4 - 2 = 2

By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]

Sn = n/2 [2(2) + (n -1) 2]

Sn = n/2  [4  + 2n – 2]

Sn = n/2 × 2 [2  + n – 1]

Sn = n [n + 1] …………(1)

 

First n Odd natural numbers are 1, 3 , 5, 7 ……….(2n - 1) which forms an AP.

Here, a = 1, d = 3 - 1 = 2

By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]

Sn = n/2 [2(1) + (n -1) 2]

Sn = n/2  [2  + 2n – 2]

Sn = n/2 × 2 [1  + n – 1]

Sn = n [ n ]

Sn = n² ………..(2)

A.T.Q

Sum of first n even numbers = k  × Sum of first n odd numbers

n (n + 1) = k x n²

[From eq 1 and 2]

n + 1 = kn²/n

k n = n + 1

k = (n + 1)/n  

Hence, the value of k is (n + 1)/n .

HOPE THIS ANSWER WILL HELP YOU….

 

Answer 2

Answer is in the attachment

Correct option: (d)