-1
$\begingroup$

I'm not sure how to go around this one. Factorizing doesn't seem to work and there isn't a clear pattern to work by that I see.

EDIT: So apparently I need to add context and stuff. I removed the checkmark from the answer because the answer for this question is one of these options (according to my book)
A:$2^{1007}$
B:$-2^{1007}$
C:$-2^{1007}-1$
D:$1$
E:$-1$

Using the formula mentioned in the answers comments: $$S=(x-x^{n+1})/(1-x)$$

I get

$$(i-1-(i-1)^{2014})/(2-i)$$

Writing $(i-1)^{2014}$ in its polar form and calculating it gives me $(i-1)^{2014}=i2^{1007}$

Further simplifying this, I get the real part as something like $(2^{1007}-3)/5$

Which is not one of the options I have.

$\endgroup$
5
  • 7
    $\begingroup$ Looks like a geometric series! $\endgroup$
    – Simon S
    Apr 19, 2015 at 21:37
  • 1
    $\begingroup$ "Factorizing doesn't seem to work" Please explain what "Factorizing" refers to in this context. $\endgroup$
    – Did
    Apr 19, 2015 at 21:40
  • $\begingroup$ Welp, not really sure what to do with that? $\endgroup$
    – John Doe
    Apr 19, 2015 at 21:40
  • $\begingroup$ $a + a^2 + \cdots + a^n = ...$ $\endgroup$
    – Simon S
    Apr 19, 2015 at 21:40
  • $\begingroup$ Basically, I tried taking out (i-1) from the successive powers of (i-1), but that didn't simplify anything (that I could see) $\endgroup$
    – John Doe
    Apr 19, 2015 at 21:41

2 Answers 2

7
$\begingroup$

This is a geometric series with first term $a=i-1$ and common ratio $r=i-1$. As such, the sum to the 2013th term is:

$$S_{2013}={a(1-r^{2013})\over 1-r}={(i-1)\left(1-(i-1)^{2013}\right)\over 1-(i-1)}={(i-1)\left(1-(i-1)^{2013}\right)\over 2-i}$$ The trickiest part is probably the 2013th power, but we can simplify it using the polar form:

$$(i-1)^{2013}=(\sqrt 2 e^{-i\pi/4})^{2013}=2^{2013/2}e^{-2013i\pi/4}=2^{2012/2}2^{1/2}e^{-2012i\pi/4}e^{-i\pi/4}=2^{1006}2^{1/2}e^{-503i\pi}e^{-i\pi/4}=-2^{1006}\sqrt 2 e^{-i\pi/4}=-2^{1006}(i-1)=2^{1006}(1-i)$$

I'll leave the rest as an exercise!

$\endgroup$
7
  • $\begingroup$ Hmm, is there any way to calculate $(i-1)^{2013}$ without using the polar form of complex numbers (or at least $e$)? I really don't get the downvotes, I genuinely didn't know this formula for geometric series. I found the exercise in a book meant for students who haven't learned these concepts yet. Either way, thanks a lot! $\endgroup$
    – John Doe
    Apr 19, 2015 at 21:54
  • $\begingroup$ you could use the binomial expansion formula for $(a+b)^n$, but it would take you a while to get back to the final result $\endgroup$
    – danimal
    Apr 19, 2015 at 21:56
  • $\begingroup$ for the geometric series sum formula, call the sum $S=x+x^2+x^3+...+x^n$. Then $xS=x^2+x^3+x^4+...+x^{n+1}$. Subtract one from the other and factorise to get $S(1-x)=x-x^{n+1}=x(1-x^n)$ so that $S={x(1-x^n)\over 1-x}$ $\endgroup$
    – danimal
    Apr 19, 2015 at 22:00
  • $\begingroup$ @Essam, I can't really understand what you have written but it looks like you are using the infinite series formula, but this question is about a finite series $\endgroup$
    – danimal
    Apr 19, 2015 at 22:11
  • $\begingroup$ wolframalpha.com/input/… which agrees - not sure what the problem is here $\endgroup$
    – danimal
    Apr 19, 2015 at 22:15
1
$\begingroup$

Since $i-1=\sqrt2\,e^{3\pi i/4}$, we have $$ \begin{align} (i-1)^{2014} &=2^{1007}\,e^{\pi i/2}\\ &=2^{1007}i \end{align} $$ Therefore, $$ \begin{align} \sum_{k=1}^{2013}(i-1)^k &=\frac{(i-1)-(i-1)^{2014}}{1-(i-1)}\\ &=\frac{(i-1)-2^{1007}i}{2-i}\\ &=\frac{-1-(2^{1007}-1)i}{2-i}\\ &=\frac{(-1-(2^{1007}-1)i)(2+i)}5\\ &=\frac{(2^{1007}-3)-(2^{1008}-1)i}5\\ \end{align} $$ Thus, the value for the real part in the question is correct. (Mathematica agrees.)

$\endgroup$

You must log in to answer this question.

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