User:Dmcq/Sandbox hide

Present version from Mathematical induction article
Algebra will now establish that


 * $$\frac{n(n + 1)}{2} + (n+1) = \frac{(n+1)((n+1) + 1)}{2}\,,$$

thereby showing that indeed P(n + 1) holds.

Script version
Algebra will now establish that


 * $$\frac{n(n + 1)}{2} + (n+1) = \frac{(n+1)((n+1) + 1)}{2}\,,$$

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

thereby showing that indeed P(n + 1) holds.

Footnoted version
Algebra will now establish that


 * $$\frac{n(n + 1)}{2} + (n+1) = \frac{(n+1)((n+1) + 1)}{2}\,,$$

thereby showing that indeed P(n + 1) holds.

Alternative footnoted version
Algebra will now establish that


 * $$\frac{n(n + 1)}{2} + (n+1) = \frac{(n+1)((n+1) + 1)}{2}\,,$$

thereby showing that indeed P(n + 1) holds.

Interpolated material
 x2 - 4

Material with references
In 370 BC, Plato’s dialog Parmenides may have contained the first inductive proof ever. Other implicit traces of mathematical induction can be found in Euclid's proof that the number of primes is infinite and in Bhaskara's "cyclic method". Other examples of inductive arguments have been found in other cultures.

Derivations
