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

Asked by vaishnavi | 29th Aug, 2012, 03:03: PM

Expert Answer:

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.

Answered by  | 29th Aug, 2012, 03:41: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.