Question 6221
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 The 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

 This completes the proof.

 Kenny