Show that the sum of first n even numbers is equal to ( 1+1/n) times the sum of first n odd numbers
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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.
Step-by-step explanation:
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