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I am trying to simplify $$ \frac{3(5^{k+1} - 1) + 4(3 \;\cdot\;5^{k+1})}{4}$$ to get to $$\frac{3(5^{k+2}-1)}{4} $$

I start with $$ \frac{3(5^{k+1} - 1) + 12\;\cdot\;5^{k+1}}{4}$$

$$= \frac{3\;\cdot\;5^{k+1} - 3 + 12\;\cdot\;5^{k+1}}{4} $$

then I am stuck.

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1  
Factor out a 3 on the numerator. – Jed Mar 8 at 1:33
up vote 11 down vote accepted

$$3(5^{k+1} - 1) + 4(3 \cdot 5^{k+1}) = \\ 3 \cdot 5^{k+1} - 3 + 12 \cdot 5^{k+1} = \\ 15 \cdot 5^{k+1} - 3 = \\ 3 \cdot 5 \cdot 5^{k+1} - 3 = \\ 3 (5 \cdot 5^{k+1} - 1) = \\ 3 (5^{k+2} - 1).$$

In going from the first line to the second line, I distribute the $3$ through on the first term in the sum: $3(5^{k+1} - 1) = 3 \cdot 5^{k+1} - 3.$ This is the first two terms in the sum on the second line. Then, I multiply through by $4$ on the second term to complete it: $4(3 \cdot 5^{k+1}) = 12 \cdot 5^{k+1}.$

Here is the same set of equations, substituting $5^{k+1}$ with $X$ until the very end:

$$3(5^{k+1} - 1) + 4(3 \cdot 5^{k+1}) = \\ 3(X - 1) + 4(3 \cdot X) = \\ 3X - 3 + 12X = \\ 15X - 3 = \\ 3 \cdot 5X - 3 = \\ 3 (5X - 1) = \\ 3 (5 \cdot 5^{k+1} - 1) = \\ 3 (5^{k+2} - 1).$$

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How did you get from $3(5^{k+1} - 1) + 4(3 \cdot 5^{k+1}) $ to $ \\ 15 \cdot 5^{k+1} - 3 $ ? – Out Of Bounds Mar 8 at 1:40
    
I added a line at the beginning to make the factoring more explicit. – John Mar 8 at 1:42
    
I expanded $5^{k+1}$ in the second step to $ 5^k \cdot 5$ then multiplied 5 by 3 and I got $ 15 \cdot 5^{k} - 3$. What happened to $ 12 \cdot 5^{k+1}$ ? How did that expression disappear ? And why $ 15 \cdot 5^{k+1} - 3$ and not $ 15 \cdot 5^k - 3$ ? – Out Of Bounds Mar 8 at 1:49
1  
I added some more explanation on the first to second line. Also, I did the whole thing again substituting $X = 5^{k+1}$; maybe that will make things clearer. – John Mar 8 at 2:05

Continuing from where you left off:

$$ \frac{3\;\cdot\;5^{k+1} - 3 + 12\;\cdot\;5^{k+1}}{4} $$

Considering the numerator:

$$3\;\cdot\;5^{k+1} + 12\;\cdot\;5^{k+1} -3$$

$$5^{k+1}(3+12)-3=$$

$$5^{k+1}(15)-3=$$

$$5^{k+1}(5.3)-3=$$

$$5^{k+2}(3)-3=$$

$$3(5^{k+2}-1)$$

So your original expressing is:

$$\frac{3(5^{k+2}-1)}{4}$$

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$$\begin{align} & \frac{3(5^{k+1} - 1) + 4(3 \;\cdot\;5^{k+1})}{4}\\ = & \frac{3(5^{k+1} - 1) + 3(4 \;\cdot\;5^{k+1})}{4}\\ = & \frac{3(1\cdot5^{k+1} + 4 \;\cdot\;5^{k+1} - 1)}{4}\\ = & \frac{3(5\cdot5^{k+1} - 1)}{4}\\ = & \frac{3(5^{k+2} - 1)}{4} \end{align}$$

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