3
$\begingroup$

I start with simplifying: $$\sum_{i=0}^n(-1)^i(\frac{1}{2})^i=\sum_{i=0}^n(-\frac{1}{2})^i$$

then:

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

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

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

am I on the right track?

$\endgroup$
3
  • 1
    $\begingroup$ That all looks fine, although both numerator and denominator are less than $1$, so you can swap the signs write it as a positive quotient. $\endgroup$
    – Chappers
    Commented Apr 19, 2015 at 20:14
  • $\begingroup$ @Chappers: Thanks for that, is it better now? $\endgroup$
    – lucidgold
    Commented Apr 19, 2015 at 20:19
  • 1
    $\begingroup$ Hang on, you've lost a $-$. Okay, now you can't really do any better, unless you want to mess around with writing it as a quotient of integers. $\endgroup$
    – Chappers
    Commented Apr 19, 2015 at 20:25

2 Answers 2

1
$\begingroup$

Here is the complete answer:

$$\sum_{i=0}^n(-1)^i(\frac{1}{2})^i=\sum_{i=0}^n(-\frac{1}{2})^i$$

then:

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

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

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

divide both sides by $(-\frac{1}{2}-1)$, we have:

$$S = \frac{(-\frac{1}{2})^{n+1} - 1}{(-\frac{1}{2}-1)}=\frac{(-\frac{1}{2})^{n+1} - 1}{(-1.5)}=\frac{1-(-\frac{1}{2})^{n+1}}{1.5}$$

$\endgroup$
0
$\begingroup$

Here's another approach using the formula for finite geometric series \begin{align*} \sum_{i=0}^nx^n=\frac{1-x^{n+1}}{1-x}\tag{1} \end{align*}

We obtain \begin{align*} \sum_{i=0}^n(-1)^i\left(\frac{1}{2}\right)^i&=\sum_{i=0}^n\left(-\frac{1}{2}\right)^i\\ &=\frac{1-\left(-\frac{1}{2}\right)^{n+1}}{1-\left(-\frac{1}{2}\right)}\\ &=\frac{2}{3}\left(1-\left(-\frac{1}{2}\right)^{n+1}\right) \end{align*}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .