If the sum of First n even natural numbers = K times the sum of first n odd natural numbers, find k

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Solution

Dear Student,

The first n even natural numbers are

$N_{1} = {2, 4 6, 8 . . . . 2 n} T h e y a r e i n a r i t h m e t i c p r o g r e s s i o n w i t h f i r s t n u m b e r a = 2 a n d c o m m o n d i f f e r e n c e d = 2 S u m o f t h e n u m b e r s i n t h i s s e q u e n c e = \frac{n}{2} . (2.2 + (n - 1) . 2) = \frac{n . (2 n + 2)}{2} = n . (n + 1) T h e f i r s t n o d d n a t u r a l n u m b e r s a r e : N_{2} = {1, 3, 5 . . . ., 1 + (n - 1) 2} S u m o f n u m b e r s i n N_{2} = \frac{n}{2} . (2.1 + (n - 1) . 2) = \frac{n . 2 n}{2} = n^{2} G i v e n, N_{1} = K . N_{2} s o, n . (n + 1) = K . n^{2} \Rightarrow K = \frac{n + 1}{n}$

Regards,

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