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) About Me  (Show Source):
You can put this solution on YOUR website!
%282n%2B1%29%5E2-%282n%2B1%29=%282n%2B1%29%28%282n%2B1%29-1%29
%282n%2B1%29%5E2-%282n%2B1%29=%282n%2B1%29%282n%29
So then, for any value n,
2n is even.
2n%2B1 is odd.
The product of an even number and an odd number is always even.


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

%282n%2B1%29%5E2-%282n%2B1%29=2k%2B1, for some k∈Z

On the other hand,

4n%5E2%2B4n%2B1-2n-1=4n%5E2%2B2n=2%282n%5E2%2Bn%29, which is an even number. This contradicts the original hypothesis.


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

    For any integer n the number  n%5E2+-n  is even.


    Indeed,  n%5E2-n = 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  n%5E2-n  is even.


Answer by MathTherapy(10551) About Me  (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 :)
%282n+%2B+1%29%5E2+-+%282n+%2B+1%29

(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,