Prove algebraically that (2n+1)^2 -(2n+1) is an even number for all the positive integer values of n.
In order to prove this, the first step is to rewrite the (2n+1)2 as (2n+1)(2n+1). This will transform the main expression to : (2n+1)(2n+1) - (2n+1). Then expand the brackets and multiply out the "(2n+1)(2n+1)" to get 4n2 + 4n + 1 and substitute this expansion into the main expression to form : 4n2 + 4n + 1 -( 2n+1). Now multiply out the bracket and simplify remember to multiply the brackets by the negative sign it then becomes "4n2 + 4n + 1 - 2n-1". This simplifies to 4n2 +2n. Now that we have expanded and simplified the expression, we have to prove that the expression "4n2 +2n" is an even number for all positive n integers. This is done by collecting the like terms by factorising. The like term in this expression is "2n" so if we factorise 4n2 +2n we get 2n*( 2n+1). It can then be spotted that 2n is always an even number for any positive integer, and that "2n+1" adds one to every even number, therefore that expression represents an odd number. Since it is known that for any value "even number * odd number = even number", therefore we can prove that (2n+1)2 -(2n+1) is therefore an even number for all positive integer values.
