The sum of first N Natural numbers is .... N * (N+1) / 2
Prove it.
So let's assume Sum as
Sum = 1 + 2 + 3 + ......N-2+N-1+N
Now Reverse it.
Sum = N+N-1+N-2+.....3+2+1
Add these.
Sum = 1 + 2 + 3 + ......N-2+N-1+N
Sum = N+N-1+N-2+.....3+2+1
2Sum = (N+1) + (N+1) + (N+1) + .....(N+1)
N+1 occurs N times in the right side , so...
2Sum = N * (N+1)
Sum = N * (N+1) /2
Hence proved.