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I have the formula $y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$, and I should work to $y = \frac{1}{2}\cdot z \cdot (z + 1)$.

Somebody showed me how it's done:

$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cdot ((z + 1) \cdot (z + 2))$
$y + (z + 1) = \frac{1}{2} \cdot (z^2 + 3z + 2)$
$y + (z + 1) = \frac{1}{2}(z^2) + \frac{1}{2}(3z) + \frac{1}{2}(2)$
$y + (z + 1) = \frac{1}{2}(z^2) + 1\frac{1}{2}z + 1$
$y = \frac{1}{2}(z^2) + 1\frac{1}{2}z + 1$ - z - 1
$y = \frac{1}{2}(z^2) + \frac{1}{2}z$
$y = \frac{1}{2}z(z + 1)$

Great! But, my try went completely wrong, and I don't understand what I'm doing wrong:

$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cdot ((z + 1) \cdot (z + 2))$
$y + (z + 1) = \frac{1}{2} \cdot (z^2 + 3z + 2)$
$y = \frac{1}{2} \cdot z^2 + 3z + 2 - z - 1$
$y = \frac{1}{2} \cdot z^2 + 2z + 1$
$y = \frac{1}{2} \cdot (z^2 + 2z + 1)$
$y = \frac{1}{2}(z^2) + \frac{1}{2}(2z) + \frac{1}{2}1$
$y = \frac{1}{2}(z^2) + z + \frac{1}{2}$

But from this last step, I can't get anywhere near $y = \frac{1}{2}z(z + 1)$, and I do not understand what I did wrong.

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    $\begingroup$ For some reason you stopped bracketing the halved terms. Ignoring the fact that this means what you've written suggests they're no longer halved, it led you to erroneously subtract the LHS without first doubling. This left you with the constant term, and an extra $z$. It can be seen that this is the only difference between yours and the correct solution. $\endgroup$
    – OJFord
    Commented Jun 3, 2014 at 1:48
  • $\begingroup$ Completely true, and in hindsight; how could I have missed it? $\endgroup$ Commented Jun 3, 2014 at 17:28

4 Answers 4

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$$y + (z + 1) = \frac{1}{2} (z + 1) (z + 2)$$ $$2y+2(z+1)=(z+1)(z+2)$$ $$2y=(z+1)(z+2)-2(z+1)$$ $$2y=(z+1)(z+2-2)$$ $$2y=z(z+1)$$ $$y=\frac{1}{2}z(z+1)$$


Where is your error? From $$y + (z + 1) = \frac{1}{2} (z^2 + 3z + 2)$$ you can say $$y+(z+1)=\frac{1}{2}z^2+\frac{3}{2}z+1$$ where instead you only multiplied $z^2$ by $1/2$.

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Your step $y=\frac12 z^2+3z+2-z-1$ is wrong because $k(a+b)=ka+kb\neq ka+b$. A faster way to tackle the question is this.

$y+z+1=\frac12 (z+1)(z+2)$

$y=\frac12 (z+1)(z+2)-(z+1)$

$y=(z+1)(\frac 12 z+1-1)$

$y=\frac12 z(z+1)$

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In your step 4, you forgot to divide (3z+2) by 2

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There is an error in the fourth line of your second set of equations:- $$y = \frac{1}{2} \cdot z^2 + 3z + 2 - z - 1$$ should be $$y = \frac{1}{2} \cdot (z^2 + 3z + 2 \color{red}{- 2z - 2})\\\Rightarrow y = \frac{1}{2} \cdot (z^2 + z)=\frac{1}{2}\cdot z(z+1)$$

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