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

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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.