Question

. Show that sum of first n even natural numbers is equal to (n+1/n)times the sum of first n odd natural numbers.​

Answer 1

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Answer 2

Answer:

Sum(n) of even numbers = n + 1/n odd numbers

Step-by-step explanation:

Sum of first n even natural numbers

Se= 2+ 4+...+ 2n =2(1+2+3+..+n) = 2* n(n+1)/2 = n(n+1)

The sum of the first N odd natural numbers

So= 1+3 +...+ (2n-1) [ arthemic series with common difference 2 ]

= n(1+2n-1)/2 = n^2.

We see that (1+ 1/n)So = (1+ 1/n)n^2 = n^2 + n = n(n+1) =Se

Hence, proved.

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