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How do I simplify this? This is the LHS of the equation and I need it to equal the RHS, which is $2-1/k+!$

$$2−\frac{k^2+k+1}{k(k+1)2}$$

Matthew Conroy
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Lil
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  • You should reduce to same denominator. – zozoens Jan 30 '14 at 13:16
  • how do I go about this since k^2+k+1 is not factorable? – Lil Jan 30 '14 at 13:17
  • Just say that $2 = \frac{2k(k+1)}{k(k+1)}$, and simplify what you get in the numerator (where you should be able to factor out $k$). – zozoens Jan 30 '14 at 13:18
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    Could you improve the formatting? The denominator of the LHS looks odd, and I have no idea what the RHS is what does "+!" mean? – MPW Jan 30 '14 at 13:27
  • http://math.stackexchange.com/questions/656791/help-with-the-algebra-in-for-this-number-theory-proof?noredirect=1#comment1382262_656791

    the second to last line in the first users answer. I don't understand how they went from the second to last line to the solution

     – Lil Jan 30 '14 at 13:29
  • That is not an equality, but an inequality. It just uses the fact that $k^2+k+1>k^2+k=k(k+1)$. – zozoens Jan 30 '14 at 13:33

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What was written in the question you reference is $$2-\{\frac{k^2+k+1}{k(k+1)^2}\} \leq 2-\frac{k(k+1)}{k(k+1)^2}$$ Note that it is not an equality. The numerator on the right expands to $k^2+k$, so the quantity being subtracted on the left is larger than the quantity being subtracted on the left. This justifies the $\le$ sign, which could have been $\lt$

Ross Millikan
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