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

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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 = n/2 [2a + (n-1) d] So, for n terms, S_n = n/2 [ 2(1) + (n-1)2] = n/2 [ 2 + 2n - 2] =n/2 (2n) = n2 ..................(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 n terms, `S_n = n/2 [ 2 (2) + (n - 1) 2] = n/2 [4 + 2n - 2 ] = n /2 ( 2 + 2n)` Solving further, we get = n ( 1 + n) = n2 + 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 n2 + n = kn2 k =(n^2 + n )/(n^2) k = (n(1 +n))/(n^2) k = ( n+ 1)/n Therefore, k = ( n+ 1)/n

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