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Solve $$(i+i^2+i^3+...+i^{102})^2$$

I have tried 2 approaches:

  1. to look at power of $2 (=-1)$ and power of $4 (=1)$ and there are $102$ elements which is even number, but because there are $4$ "types" $\{i,-1,-i,1\}$ it did not work it out
  2. to look at $$(i+i^2+i^3+...+i^{102})^2=\left(\sum_{k=1}^{102}i^k\right)^2$$ but $$\left(\sum_{k=1}^{102}i^k\right)^2\neq \sum_{k=1}^{102}i^{2k}$$ so it was not useful.

Any suggestion how to tackle this?

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  • $\begingroup$ WolfRamAlpha says -2i $\endgroup$ – Eric Lee May 25 '17 at 13:45
  • $\begingroup$ Are you familiar with the sum of a geometric progession $\sum_{k=1}^{N} r^k = \frac{r-r^{N+1}}{1-r}$ $\endgroup$ – sharding4 May 25 '17 at 13:45
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HINTS:

$$i+i^2+i^3+i^4=0,$$

$$i^{4m+n}=i^{n}.$$

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    $\begingroup$ It may be useful for the OP to note of course a slight generalisation; that if $\omega$ is an $n^{\text{th}}$ root of $1$; so $\omega^n -1 =0$, then $\omega^n-1= (\omega -1)(\omega^{n-1} + \dots + \omega + 1)$. So if $\omega \ne 1$ then we get that $\omega^{n-1} + \dots + \omega + 1=0$ $\endgroup$ – SEWillB May 25 '17 at 14:13
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One trick for this kind of problem is to work it out for small numbers and see if you can find a pattern. Try working out: $$ i + i^2 $$ $$ i + i^2 + i^3 $$ $$ i + i^2 + i^3 + i^4 $$ $$ i + i^2 + i^3 + i^4 + i^5 $$ $$ i + i^2 + i^3 + i^4 + i^5 + i^6$$ Do you see a pattern?

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  • $\begingroup$ Should the $i^3$'s on the left toward the bottom be squared instead? $\endgroup$ – Chickenmancer May 25 '17 at 15:00
  • $\begingroup$ @Chickenmancer Hah, yes, thank you $\endgroup$ – Neal May 25 '17 at 15:09
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$$a + a^2 + a^3 + \dots a^n = a\frac{1-a^{n}}{1-a}$$

Therefore,

\begin{align*} (i+i^2 + \cdots + i^{102}) &= i \frac{1-i^{102}}{1-i}\\ & = i \frac{1-i^{100}i^2}{1-i} \\ &= i \frac{1+(i^{4})^{25}}{1-i}\\ &= \frac{2i}{1-i}\\ &= -1 + i \end{align*}

Squaring yields $-2i$.

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    $\begingroup$ I don't think geometric summing is useful here. The trick is that most of the sum cancels to zero. $\endgroup$ – Neal May 25 '17 at 13:48
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    $\begingroup$ Because in this case, the seeing that the terms are periodic allows a more elegant solution. (Not saying you're wrong, just saying that this solution obscures the fact that the summands cancel) $\endgroup$ – Neal May 25 '17 at 13:52
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    $\begingroup$ Just giving an alternative method. No need to downvote. $\endgroup$ – user370967 May 25 '17 at 13:53
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    $\begingroup$ Couldn't disagree more with you "geometric progression haters." You can see the cancellation which took place in the rear view mirror after you have quickly found the answer using the sum of a geometric progression. $\endgroup$ – sharding4 May 25 '17 at 13:54
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    $\begingroup$ @sharding4 How do I find $i^{103}$ and how does that show me that the sum of four consecutive powers of $i$ is zero? I don't hate GP --- it is very useful --- but in this case I prefer other approaches. $\endgroup$ – JP McCarthy May 25 '17 at 13:56
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Your first idea works:

If $n=4k+1$ then $i^n = i$,

If $n=4k+2$ then $i^n = -1$,

If $n=4k+3$ then $i^n = -i$,

If $n=4k$ then $i^n = 1$

So, $i+i^2+i^3+i^4=0$ $i^5+i^6+i^7+i^8=0$

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.

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$i^{97}+i^{98}+i^{99}+i^{100}=0$

Then, in the sum $i+i^2+i^3+...+i^{102}$ the only left terms are $i^{101}+i^{102}=i-1$

Thus, $(i+i^2+i^3+...+i^{102})^2=(i-1)^2= i^2 -2i+1=-1-2i+1=-2i$

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$$i=i\\i^2 =-1\\i^3 =-i \\i^4 =1$$ Thus, $$\sum_{k=1}^{4n} i^k =0\ \text{for positive integer}\ n$$, Hence, $$\sum_{k=1}^{102} i^k=\sum_{k=1}^{2} i^k=-1+i$$ And it's square equals $-2i$

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