Proving algebraic operations for this simple problem

I cannot wrap my heard around solving this:

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

\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?

• In the second line $\mu\frac{1-(1/2)^{n+1}}{1/2}=\mu\frac{1}{1/2}-\mu\frac{(1/2)^n(1/2)}{1/2}$. – user276115 Nov 25 '16 at 11:27

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

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

thanks guys!

\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

• In the first step, it should be: $\mu\frac{1-(\frac{1}{4})^{n+1}}{\frac{3}{4}}=\mu\frac{1}{\frac{3}{4}}-\mu\frac{(\frac{1}{4})^{n}(\frac{1}{4})^{1}}{\frac{3}{4}}$ then it follows $\mu\frac{4}{3}-\mu\frac{4}{3}{(\frac{1}{4})^{n}(\frac{1}{4})^{1}} = \frac{4}{3}\mu[1-(\frac{1}{4})^{n+1}]$. – Simon Woo Nov 28 '16 at 8:04