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I'm looking at a solution to a math problem and there are obviously some rules regarding multiplication of fractions that I don't know.

Can someone make any sense of this?

$$s_n = 625 \cdot \frac{\left(\frac 1 5\right)^n - 1}{\frac 1 5 - 1} = 625 \cdot \frac{\left(\frac 1 5\right)^n - 1}{- \frac 4 5} = - \frac{3125}{4} \cdot \left(\left(\frac 1 5\right)^n -1\right) = \frac{3125}{4} \cdot \left( 1 - \left(\frac 1 5\right)^n\right)$$

Can someone explain to me how you get from each step to another?

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    $\begingroup$ Which part do you not understand? What do you know about multiplication and division of fractions? $\endgroup$ – Martin R Apr 20 '15 at 5:14
  • $\begingroup$ I get step 1>step 2, that one's easy. It's how you get from step 2 to 3 and 3 to 4 that I don't understand. What is done with 625 and -4/5 to get -3125/4? $\endgroup$ – Julian Nikolay Krogh-Fredrikse Apr 20 '15 at 5:35
  • $\begingroup$ Note that $a\cdot \dfrac{1}{\frac{2}{c}}=a\cdot \frac{c}{2}$. That's the principle being used there. $\endgroup$ – Sujaan Kunalan Apr 20 '15 at 5:39
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First step: subtraction

Second step: $\frac{a}{\frac{b}{c}}=\frac{c}{b}a$

Third step: $(-c)(a-b)=(-1)c(a-b)=c(-1)(a-b)=c(-a-(-b))=c(b-a)$

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The denominator $\frac{1}{5}-1 = \frac{1}{5}-\frac{5}{5}=\frac{1-5}{5}=\frac{-4}{5}$

Denominator's denominator is numerator, that makes $625 \cdot \frac{()}{-\frac{4}{5}}=-\frac{625 \times 5\times ()}{4} = -\frac{3125\times ()} {4}$

Lastly,$-\frac{3125}{4}=\frac{3125 \times \color{red}{-1}}{4}$ and when this $\color{red}{-1}$ taken inside the bracket $\left(\color{red}{-1} \times (\frac{1}{5})^n \color{red}{-1} \times -1\right)=\left( 1 - (\frac{1}{5})^n\right)$ because negative number multiplied by negative number becomes positive and positive number number multiplied by negative number becomes negative.

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I'll try to lay out what happens in each step (equality).

Step 1. $\frac 1 5 - 1$ turns into $- \frac 4 5$. I assume you don't have a problem with this: it's just subtraction. Note that $1 = \frac 5 5$, so $\frac 1 5 - 1 = \frac 1 5 - \frac 5 5 = \frac{-4}{5} = - \frac 4 5$.

Step 2. The author does two things at once here: they multiply the fraction by $625$, and they carry out the division by $- \frac 4 5$. That might be confusing, so let's review what happens here. The first one is fraction multiplication. The big rule is: $$\frac a b \cdot \frac c d = \frac {ac}{bd}$$ Here's a good trick: $$a = \frac a 1$$ And this little rule follows from the trick and the big rule: $$b \cdot \frac 1 a = \frac b a = \frac 1 a \cdot b$$ The second topic, a little more complicated, concerns fraction division (the first equality here is just to be clear on notation): $$\frac {\frac a b} {\frac c d} = \frac{a/b}{c/d} = \frac a b \cdot \frac d c = \frac {ad}{bc}$$ One important thing about negation (this makes sense if you consider a negative sign to be the same as multiplying by $(-1)$): $$\frac {-a} b = - \frac a b = \frac a {-b}$$ So let's go through the equality slowly. We'll do the division part first.

$$625 \cdot \frac{\left(\frac 1 5\right)^n -1}{ - \frac 4 5} = 625 \cdot \frac{\frac{\left(\frac 1 5\right)^n -1}1}{ - \frac 4 5} = 625 \cdot \frac{\frac{\left(\frac 1 5\right)^n -1}1}{ \frac {4} {-5}} = 625 \cdot \frac{\left(\frac 1 5\right)^n -1} 1 \cdot {\frac {-5} 4}$$ Continuing, $$625 \cdot \frac{\left(\frac 1 5\right)^n -1} 1 \cdot {\frac {-5} 4} = 625 \cdot \frac{\left(\left(\frac 1 5\right)^n -1\right)\cdot(-5)} {1 \cdot 4} = 625 \cdot \frac{\left(\left(\frac 1 5\right)^n -1\right)\cdot(-5)} {4}$$

Then we rearrange: $$ 625 \cdot \frac{\left(\left(\frac 1 5\right)^n -1\right)\cdot(-5)} {4} = 625 \cdot \frac{(-5)} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right) = \frac{625 \cdot(-5)} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right)$$ Multiply... $$\frac{625 \cdot(-5)} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right) = \frac{-3125} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right) = - \frac{3125} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right)$$

Which completes that step.

Now the next step: we note that $$ - \frac a b = \frac {-a} b = \frac{(-1) \cdot a}{b} = (-1) \cdot \frac a b$$

So we get: $$- \frac{3125} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right) = (-1) \cdot \frac{3125} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right)$$

And then we multiply the $(-1)$ with the right part of the equation. We can do this because: $$a \cdot b \cdot c = ab \cdot c = a \cdot bc = b \cdot ac = b \cdot a \cdot c$$ Et cetera. The point being is that when we multiply, the exact order in which we multiply the elements doesn't matter. (This is called commutativity.)

And now note that $(-1) \cdot (a-b) = b-a$. This makes sense because $(a-b) = (a + (-b))$, and $(-1) \cdot (a+(-b)) = ((-1) \cdot a) + ((-1) \cdot (-b) = ((-a) + (b)) = (b-a)$. Remember that $(-1) \cdot (-1) = 1$. So we get:

$$(-1) \cdot \frac{3125} {4} \cdot \left(\left(\frac 1 5\right)^n -1\right) = \frac{3125} {4} \cdot (-1) \cdot \left(\left(\frac 1 5\right)^n -1\right)$$

And
$$\frac{3125} {4} \cdot (-1) \cdot \left(\left(\frac 1 5\right)^n -1\right) = \frac{3125} {4} \cdot (-1) \cdot \left(\left(\frac 1 5\right)^n + (-1)\right) $$ Now multiplying... $$ \frac{3125} {4} \cdot (-1) \cdot \left(\left(\frac 1 5\right)^n + (-1)\right) = \frac{3125} {4} \cdot \left(\left( (-1) \cdot \left(\frac 1 5\right)^n\right) + ((-1) \cdot(-1))\right) $$ Finally, we're almost there... $$ \frac{3125} {4} \cdot \left(\left( (-1) \cdot \left(\frac 1 5\right)^n\right) + ((-1) \cdot(-1))\right) = \frac{3125} {4} \cdot \left(-\left(\frac 1 5\right)^n + 1\right) $$ And of course: $$ \frac{3125} {4} \cdot \left(-\left(\frac 1 5\right)^n + 1\right) = \frac{3125} {4} \cdot \left(1-\left(\frac 1 5\right)^n\right) $$

As desired.

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  • $\begingroup$ This ended up being a little longer than intended, but at least it was thorough. $\endgroup$ – Newb Apr 20 '15 at 5:34
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Maybe you are confused about the multiplication of $\frac 1{5^n}$ with $-1$?

From first to second just substract $1$ from $\frac 15$.

Second part is to multiply $625$ with $\frac{-5}4$, which comes from $1/(-4/5)$ by dividing,

Finally we take $-1$ into the $(\frac 1 {5^n} - 1)$, and since every negative number will turn to positive and every positive number will turn to negative, each number will change its sign. In order to show that $n$ doesn't matter for our case,

for $n=1$: $-1 \cdot(1-1/5) = -4/5 = (1/5-1)$,

for $n=2$: $-1 \cdot (1-1/25) = -24/25 = (1/25-1)$,

As you can see, it doesn't matter whether n is odd or even we just do the simple multiplication.

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