Proof by induction

One proof of triangular numbers is by induction.

T(n) = 1 + 2 + 3 + ...+ n = [n ( n+ 1)]/ 2

Proof: Let n = 1.

If n = 1, then [1 (2)] / 2 = 1, which is true.

Let n = k such that k is a positive integer

1 + 2 + 3 + ... + k = [k(k+1)]/2

I must show that k + 1 is true.

1 + 2 + 3 + ... + k + (k + 1) = (1 + 2 + 3 + ... + k ) + (k + 1)

...............................................= [k(k+1)]/2 + (k + 1)

...............................................=[k(k+1)]/2 = [2(k+1)]/2

..............................................=[k^2 + 3k + 3]/2

..............................................=[(k + 1) (k + 2)] /2

..............................................= [(k + 1) (k + 1 + 1)] /2

So by mathematical induction, this statement is true for all n. Q.E.D.

Proof by Numerical

t (n) = 1 + 2 + 3 + .....+ n -1 + n

t(n) = n + n-1 + n-2 + ...+2 + 1

2 t(n) = (n + 1) + (n + 1) + (n + 1) +...+ (n+1)

2 t(n) = n ( n+1)

t(n) = [n ( n+ 1)]/ 2

Geometric Proof

1 x 1 1

2 x 3

1 + 2

4 x 3

1 + 2 + 3

5 x 4

1 + 2 + 3 + 4

So my conclusion is that

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

1 + 2 + 3 + ...+ n