Apparently $\displaystyle \frac{1}{28}(2x-1)^7 + \frac{1}{24}(2x-1)^6 = \frac{1}{168}(2x-1)^6(12x+1)$ I can't see it, I'd appreciate some help.
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Originally Posted by JeWiSh Apparently $\displaystyle \frac{1}{28}(2x-1)^7 + \frac{1}{24}(2x-1)^6 = \frac{1}{168}(2x-1)^6(12x+1)$ I can't see it, I'd appreciate some help. Take out a common factor of $\displaystyle (2x - 1)^6$. This gives $\displaystyle (2x - 1)^6\left[\frac{1}{28}(2x - 1) + \frac{1}{24}\right]$ Can you go from there?
Originally Posted by Prove It Take out a common factor of $\displaystyle (2x - 1)^6$. This gives $\displaystyle (2x - 1)^6\left[\frac{1}{28}(2x - 1) + \frac{1}{24}\right]$ Can you go from there? Ahhh, yes got it, thanks.
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