SHOW THAT THE SUM OF THE FIRST N EVEN NATURAL NO. IS EQUAL TO (1+1/N) TIMES THE SUM OF THE FIRST N ODD NATURAL NO.

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)