0
$\begingroup$

Given $$i+i^2+i^3+...+i^{2012}$$ find value of this. I do not know how to find value given so much numbers.I would like any hint given to solve these kind of problems.

$\endgroup$
  • $\begingroup$ Hint: $\;1+x+x^2+\cdots+x^{2011} = (1-x^{2012}) / (1-x)$ $\endgroup$ – dxiv Jun 22 '17 at 17:38
  • $\begingroup$ Add some of the first terms together. Can you find a pattern? Always try to find a pattern in scary questions like these. $\endgroup$ – greenturtle3141 Jun 22 '17 at 17:42
  • $\begingroup$ @greenturtle3141 since the first 4 terms equal 0 , the answer will be 0 since the values are repeated $\endgroup$ – trying to learn Jun 22 '17 at 17:48
2
$\begingroup$

Hint:

$$i+i^2+i^3+i^4=i+(-1)+(-i)+1=0$$

Try to figure out what is

$$i^5+i^6+i^7+i^8.$$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Well since i^5=i, i^6=-1 and so on , the answer will be 0 I guess $\endgroup$ – trying to learn Jun 22 '17 at 17:45
  • $\begingroup$ yes, answer is $0$. Try to use geometric series approach as suggested by dxiv as well. It's good to tackle a problem using multiple approaches. $\endgroup$ – Siong Thye Goh Jun 22 '17 at 17:48
0
$\begingroup$

$$i+i^2+\cdots+i^{2012}=x \stackrel{×i} \Rightarrow$$

$$i^2+i^3+\cdots+i^{2012}+i^{2013}=xi \stackrel{(1)-(2)} \Rightarrow$$

$$i-i^{2013}=x(1-i) \Rightarrow x=\frac{i-(i^2)^{1006}\cdot i}{1-i}=0.$$

| cite | improve this answer | |
$\endgroup$

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.