Prove that the $\displaystyle (2n+1)$th term of the sequence $\displaystyle U_n=n^{2}-1$ is a multiple of $\displaystyle 4$. How do I do this?
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$\displaystyle U_{2n+1}=(2n+1)^2-1=4n^2+4n+1-1=4(n^2+n)$ So it is a multiple of 4.
EDIT: Nevermind, I just realised what a fool I'm being. I'll try it again soon.
Originally Posted by Viral $\displaystyle U_{2n+1}=(2n+1)^2-1=4n^2+1-1$ That's what I keep getting =\ . The expansion of $\displaystyle (2n+1)^2$ should be $\displaystyle (2n+1)^2 = 4n^2+4n+1$
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