Hi

Shame on me for asking this but

$\displaystyle \bigg(\frac{(n+1)(n+2)}{2}\bigg)^2 - \bigg(\frac{n(n+1)}{2}\bigg)^2$

Do you multiply everything out before subtracting ?

I got the answer n+1 but it's wrong.

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- Oct 6th 2009, 03:14 AMJonessimple algebra question
Hi

Shame on me for asking this but

$\displaystyle \bigg(\frac{(n+1)(n+2)}{2}\bigg)^2 - \bigg(\frac{n(n+1)}{2}\bigg)^2$

Do you multiply everything out before subtracting ?

I got the answer n+1 but it's wrong. - Oct 6th 2009, 03:27 AMGrandad
Hello JonesNo, not if you want the easiest method!

Note that each term has a common factor $\displaystyle \frac{n+1}{2}$, so take it out (not forgetting that it has to be squared):

$\displaystyle \bigg(\frac{(n+1)(n+2)}{2}\bigg)^2 - \bigg(\frac{n(n+1)}{2}\bigg)^2= \Big(\frac{n+1}{2}\Big)^2\Big((n+2)^2-n^2\Big)$

Even now, there's no need to multiply out. Use 'difference of two squares' in the second bracket.

Grandad - Oct 6th 2009, 04:35 AMJones
Of curse!

Stupid me.

Thank you daddy!