Correct answer is D. \(\frac{n + 1}{n}\)
Given: the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers.
To find: The value of k
Solution: Sum of terms of A.P = n/2 ( 2a + (n-1) d )First n even natural numbers are: 2, 4, 6, 8, ............
It forms an AP where first term a = 2 and common difference d = 4 – 2 = 2
⇒ sum of n even natural terms = \(\frac{n}{2}\)x {2 x 2 + (n–1)2}
= \(\frac{n}{2}\)( 4 + 2n - 2 )
=\(\frac{n}{2}\)( 2 + 2n )
=\(\frac{n}{2}\) × 2 ( 1 + n )
= n (n + 1)
First n odd natural numbers are: 1, 3, 5, 7, ............
It forms an AP where first term is a = 1 and the common difference d = 3 – 1 = 2
Now sum of n terms = \(\frac{n}{2}\){2 x 1 + (n–1)2}
= \(\frac{n}{2}\)( 2 + 2n - 2 )= \(\frac{n}{2}\)(2n )
= n x n
= n2
Now, According to given condition Sum of first n even numbers = k X (Sum of first n odd numbers)
⇒ n (n + 1) = k x n2
⇒ k x n = n + 1
⇒ k = \(\frac{(n+1)}{n}\)