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. 
 A: 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:
$$\frac{3(5^{k+2}-1)}{4}$$
A: $$\begin{align}
& \frac{3(5^{k+1} - 1) + 4(3 \;\cdot\;5^{k+1})}{4}\\
= & \frac{3(5^{k+1} - 1) + 3(4 \;\cdot\;5^{k+1})}{4}\\
= & \frac{3(1\cdot5^{k+1} + 4 \;\cdot\;5^{k+1} - 1)}{4}\\
= & \frac{3(5\cdot5^{k+1} - 1)}{4}\\
= & \frac{3(5^{k+2} - 1)}{4}
\end{align}$$
A: $$3(5^{k+1} - 1) + 4(3 \cdot 5^{k+1}) = \\ 3 \cdot 5^{k+1} - 3 + 12 \cdot 5^{k+1} = \\ 15 \cdot 5^{k+1} - 3 = \\ 3 \cdot 5 \cdot 5^{k+1} - 3 = \\ 3 (5 \cdot 5^{k+1} - 1) = \\ 3 (5^{k+2} - 1).$$
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:
$$3(5^{k+1} - 1) + 4(3 \cdot 5^{k+1}) = \\ 3(X - 1) + 4(3 \cdot X) = \\ 3X - 3 + 12X = \\ 15X - 3 = \\ 3 \cdot 5X - 3 = \\ 3 (5X - 1) = \\ 3 (5 \cdot 5^{k+1} - 1) = \\ 3 (5^{k+2} - 1).$$
