1
$\begingroup$

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?

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

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 $\mu\frac{4}{3}-\mu\frac{4}{3}{(\frac{1}{4})^{n}(\frac{1}{4})^{1}} = \frac{4}{3}\mu[1-(\frac{1}{4})^{n+1}]$. $\endgroup$ – Simon Woo Nov 28 '16 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.