<s>[INST] How can I simplify this rational expression? 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. 
 [/INST] $$3(5^{k+1} - 1) + 4(3 \cdot 5^{k+1}) = \\ 3 \cdot 5^{k+1} - 3 + 12 \cdot 5^{k+1} = \\ 15 \cdot 5^{k+1} - 3 = \\ 3 \cdot 5 \cdot 5^{k+1} - 3 = \\ 3 (5 \cdot 5^{k+1} - 1) = \\ 3 (5^{k+2} - 1).$$
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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