# Sequential Proof

• Oct 4th 2009, 11:33 AM
Viral
Sequential Proof
Prove that the $(2n+1)$th term of the sequence $U_n=n^{2}-1$ is a multiple of $4$.

How do I do this?
• Oct 4th 2009, 11:58 AM
james_bond
$U_{2n+1}=(2n+1)^2-1=4n^2+4n+1-1=4(n^2+n)$ So it is a multiple of 4.
• Oct 4th 2009, 11:59 AM
Viral
EDIT: Nevermind, I just realised what a fool I'm being. I'll try it again soon.
• Oct 4th 2009, 12:00 PM
e^(i*pi)
Quote:

Originally Posted by Viral
$U_{2n+1}=(2n+1)^2-1=4n^2+1-1$

That's what I keep getting =\ .

The expansion of $(2n+1)^2$ should be

$(2n+1)^2 = 4n^2+4n+1$