Proving algebraic operations for this simple problem

Question

I cannot wrap my heard around solving this:

$$\mu\frac{1-(\frac{1}{2})^{n+1}}{\frac{1}{2}}=2\mu\left[1-\left(\frac{1}{2}\right)^{n+1}\right]$$

I have done this instead:

\begin{align} \mu\frac{1-(\frac{1}{2})^{n+1}}{\frac{1}{2}}&=\mu\frac{1}{\frac{1}{2}}-\frac{(\frac{1}{2})^{n}(\frac{1}{2})^{1}}{\frac{1}{2}}\\ &=2\mu-\frac{1}{2}^{n} \end{align}

What have I done wrong?

Answer 1

The $\mu$ should be distributed on both terms in the first step:

$$\mu\frac{1-(\frac{1}{2})^{n+1}}{\frac{1}{2}}=\mu\frac{1}{\frac{1}{2}}-\mu\frac{(\frac{1}{2})^{n}(\frac{1}{2})^{1}}{\frac{1}{2}}$$

Answer 2

thanks guys!

What about this one

\begin{align} \mu\frac{1-(\frac{1}{4})^{n+1}}{\frac{3}{4}}&=\mu\frac{1}{\frac{3}{4}}-\frac{(\frac{1}{4})^{n}(\frac{1}{4})^{1}}{\frac{3}{4}}\\ \end{align}

I don't know how to proceed from this other than

\begin{align} \frac{4}{3}\mu-\mu(\frac{3}{4})^{n+1} \end{align}

which is obviosly wrong. As the result should be

\begin{align} \frac{4}{3}\mu[1-(\frac{1}{4})^{n+1}] \end{align}

Thanks