How do I get from here to here? - basic algebra

Hi!

I have no idea how to do this because the book I'm reading doesn't show me how:

A x sum of 1/(1+r)^{t }where t goes on for however long, apparently equals A/r x [1 - 1/(1+r)^{t }]

how do you get there?

I'd be so thankful for someone to show me :))

Re: How do I get from here to here? - basic algebra

Hello, mathbeginner97!

There is a typo . . .

Quote:

Show that: .

$\displaystyle A\sum^n_{t=1} \frac{1}{(1+x)^t} \;=\;\frac{A}{x}\left[1 - \frac{1}{(1+x)^n}\right]$

We have: .$\displaystyle S \;=\; \frac{A}{1+x} + \frac{A}{(1+x)^2} + \frac{A}{(1+x)^3} + \cdots + \frac{1}{(1+x)^n}$

This is geometric series with first term $\displaystyle a \,=\,\frac{A}{1+x},$

. . common ratio $\displaystyle r \,=\,\frac{1}{1+x}$ and $\displaystyle n$ terms.

The sum is: .$\displaystyle S \;=\;a\,\frac{1-r^n}{1-r}$

$\displaystyle S \;=\;\frac{A}{1+x}\left[\frac{1-(\frac{1}{1+x})^n}{1-(\frac{1}{1+x})}\right] \;=\;\frac{A}{1+x}\left[\frac{1 - \frac{1}{(1+x)^n}}{\frac{1+x-1}{1+x}}\right] \;=\;\frac{A}{1+x}\left[\frac{1-\frac{1}{(1+x)^n}}{\frac{x}{1+x}}\right]$

$\displaystyle S \;=\;A\left[\frac{1-\frac{1}{1+x)^n}}{x}\right] \;=\;\frac{A}{x}\left[1 - \frac{1}{(1+x)^n}\right]$

Re: How do I get from here to here? - basic algebra

Thank you very much!

I appreciate your help a lot! :)