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- Jan 26, 2012

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Thread starter
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- #1

- Jan 26, 2012

- 4,055

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- #2

- Jan 26, 2012

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1) MarkFL

2) soroban

3) veronica1999

4) anemone

5) mathmaniac

6) BAdhi

This week we received solutions with very different approaches to solving the problem so I will include more solutions than usual.

Solution:

(from MarkFL)

$\displaystyle \sum_{k=1}^{23}i^k=5\left(i+i^2+i^3+i^4 \right)+\left(i+i^2+i^3 \right)=5\left(i-1-i+1 \right)+\left(i-1-i \right)=5(0)-1=-1$

(from soroban)

Then $S+1 \:=\:1 + i + i^2 + i^3 + \cdots + i^{23}$

The right side is a geometric series

. . with first term $a = 1$, common ratio $r = i$, and $n = 24$ terms.

Its sum is: .$1\cdot\dfrac{1 - i^{24}}{1-i} \:=\:\dfrac{1-1}{1-i} \:=\:0$

Therefore: .$S+1 \;=\;0 \quad\Rightarrow\quad S \:=\:-1$

(from anemone)

$\displaystyle=i+i^2+i^2.i+(i^2)^2+(i^2)^2.i+(i^2)^3+(i^2)^3.i+(i^2)^4+(i^2)^4.i+\cdots++(i^2)^{11}+(i^2)^{11}.i$

$\displaystyle=i-1-i+1+i+-1-i+1+i-\cdots-i-1$

Now, if we group the second and third terms together and fourth and fifth terms together, and so on and so forth, i.e.

$\displaystyle=i-(1+i)+(1+i)-(1+i)+(1+i)-\cdots-(i+1)$

we see that the first group and second group cancel out perfectly and this pattern continues for another total of 4 pairs of such cancel-able terms and yields the following result:

$\displaystyle=i-\cancel {(1+i)}+\cancel {(1+i)}-\cancel {(1+i)}+\cancel {(1+i)}-\cdots-\cancel {(1+i)}+\cancel {(1+i)}-(i+1)$

$\displaystyle=i-(1+i)$

$\displaystyle=-1$

(from BAdhi)

P&=\sum \limits_{k=1}^{23} i\\

&=i + \sum \limits_{k=1}^{11}i^{2k}+i^{2k+1}\\

&=i+ \sum \limits_{k=1}^{11}i^{2k}(1+i)\\

&=i+ \sum \limits_{k=1}^{11}(i^2)^k(1+i)\\

&=i+ \sum \limits_{k=1}^{11}(-1)^k\underbrace{(1+i)}_{a}\\

&=i+ \sum \limits_{k=1}^{11}(-1)^ka\\

&=i+(-1)a+\sum \limits_{k=2}^{11}(-1)^ka\\

&=i-(1+i)+\sum \limits_{k=1}^{5}(-1)^{2k}a+(-1)^{2k+1}a\\

&=-1+\sum \limits_{k=1}^{5}\underbrace{a+(-1)a}_{=0}\\

&=-1\\

\end{align*}$$

therefore,

$$i+i^2+i^3+\cdots+i^{23}=-1$$

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