SOLUTION: Prove algebraically that (2n+1)^2-(2n+1) is an even number for all positive integer values of n. Please help :)

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Question 924485: Prove algebraically that
(2n+1)^2-(2n+1) is an even number
for all positive integer values of n.
Please help :)

Found 4 solutions by Fombitz, amalm06, ikleyn, MathTherapy:
Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!


So then, for any value ,
is even.
is odd.
The product of an even number and an odd number is always even.


Answer by amalm06(224)   (Show Source): You can put this solution on YOUR website!
Suppose it's an odd number instead. That is,

=, for some k∈Z

On the other hand,

, which is an even number. This contradicts the original hypothesis.


Answer by ikleyn(52777)   (Show Source): You can put this solution on YOUR website!
.
It is easy to prove more general statement:

    For any integer n the number    is even.


    Indeed,   = n*(n-1)  is the product of two consecutive integers.

    Of the two, one integer inevitably is even.


    Therefore, the product,  n*(n-1)  is even.


    Hence, the original    is even.


Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!

Prove algebraically that
(2n+1)^2-(2n+1) is an even number
for all positive integer values of n.
Please help :)

(2n + 1)[(2n + 1) – 1]
(2n + 1)(2n + 1 – 1)
(2n + 1)(2n), or 2n(2n + 1)
2n is an EVEN NUMBER, and adding 1 to an even number creates an ODD number
Therefore,
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