Posted by **Jennie** on Sunday, March 4, 2007 at 10:24pm.

proof by mathmatical induction that the sum of the first n natural numbers is equal n(n+1)/2

It's true for n = 1. Assume that it is true for some n. Then the sum of the first n+1 natural integers can be obtained by dding the last number n+1 to n(n+1)/2. So, the assumption that it is true for n leads to the conclusion that for n+1 the sum must be:

n+1 + n(n+1)/2.

If the formula is correct for n+1 also, then this must be the same as:

(n+1)(n+2)/2.

Expanding out the last bracket gives:

(n+1)(n+2)/2 = (n+1)*n/2 + (n+1)*2/2 =

n+1 + n(n+1)/2.

Thank you Count Iblis

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