Question

Perform the following operations.

1.) [(i)(i^2)(i^3)...(i^999)] / i + i^2 +i^3 + ... + i^999

2.) (...(((i)^1)^2)^3)...)^2014

Answer 1

Is \(i\) the imaginary number \(i=\sqrt{-1}\) ?

Answer 2

@SithsAndGiggles Yes.

Answer 3

Okay, so for the first one, you have
\[\large\frac{i\times i^2\times\cdots\times i^{999}}{i+i^2+\cdots+i^{999}}\]
For the numerator, use the property that for any \(b>0\), you have
\[\large b^a\times b^c=b^{a+c}\]
which means the numerator is rewritten as
\[\large i\times i^2\times\cdots\times i^{999}=i^{1+2+\cdots+999}\]
The exponent is the sum of consecutive integers 1 to 999. Recall this useful formula for evaluating this sum:
\[\large\sum_{k=1}^nk=1+2+\cdots+n=\frac{n(n+1)}{2}\]
When \(n=999\), you have
\[\large i^{1+\cdots+999}=i^{(999(999+1))/2}=i^{499500}\]
Since 499500 is a multiple of 4, you can reduce this to
\[\large i^{499500}=\left(i^4\right)^{124875}=1^{124875}=1\]
So we've found that
\[\large\frac{i\times i^2\times\cdots\times i^{999}}{i+i^2+\cdots+i^{999}}=\frac{1}{i+i^2+\cdots+i^{999}}\]
Does this make sense so far?

Answer 4

@SithsAndGiggles Yes. How do you solve the denominator?

Answer 5

I'm back, sorry for the wait.

I see two possible ways to rewrite the denominator.

For one way, I think you can use the formula for finite geometric sums.
\[\large \sum_{i=1}^nar^{i-1}=\frac{a(1-r^n)}{1-r}\]
In this case, \(r=i\). The denominator is
\[\large \begin{align*}i^{2-1}+i^{3-1}+\cdots+i^{1000-1}&=\sum_{k=2}^{1000}i^{k-1}\\&=\sum_{k=1}^{1000}i^{k-1}-1\\&=\frac{1-i^{1000}}{1-i}-1\\
&=\frac{1-\left(i^4\right)^{250}}{1-i}-1\\
&=\frac{1-1^{250}}{1-i}-1\\
&=0-1\\
&=-1
\end{align*}\]
This means the first expression in your question reduces to \(\dfrac{1}{-1}=-1\).

The other way would have been to count how many multiples of \(i\), \(i^2\), \(i^3\), and \(i^4\) there are, but I think that takes a bit longer.

Answer 6

@SithsAndGiggles I'm sorry but I don't get the solution... I solved it earlier using the multiples and it was long, and I think your solution is shorter. How did you solve that?

Answer 7

All the steps are listed in my previous comment. If you don't know the geometric sum formula, stick to the multiples method.

Answer 8

@SithsAndGiggles Why did you put "-1" in the second step?

Answer 9

That's because the formula for the summation applies to a starting index of \(k=1\), not \(k=2\). Note that
\[\large\sum_{k=2}^{1000}i^{k-1}=i^1+i^2+\cdots+i^{999}\]
whereas
\[\large\sum_{\color{red}{k=1}}^{1000}i^{k-1}=\color{red}{i^0}+i^1+i^2+\cdots+i^{999}\]
Since \(i^0=1\), you have
\[\large\sum_{\color{red}{k=1}}^{1000}i^{k-1}=\color{red}{1}+i^1+i^2+\cdots+i^{999}\]
and so we move this 1 to the left side:
\[\large\begin{align*}\sum_{\color{red}{k=1}}^{1000}i^{k-1}-\color{red}1&=i^1+i^2+\cdots+i^{999}\\
&=\sum_{k=2}^{1000}i^{k-1}
\end{align*}\]

Answer 10

For the second question, consider the first four powers:
\[\large \left(\left(\left(\left(\left(\left(i\right)^1\right)^2\right)^3\right)^4\right)^{\cdots}\right)^{2014}\]
Recall the property of exponents:
\[\large \left(a^b\right)^c=a^{bc}=a^{cb}=\left(a^c\right)^b\]
This means we can rearrange the order of the exponents so that \(i\) is raised to the fourth power first:
\[\large \left(\left(\left(\left(\left(\left(i\right)^\color{red}4\right)^2\right)^3\right)^\color{red}1\right)^{\cdots}\right)^{2014}\]
Since \(i^4=1\), you have
\[\large \left(\left(\left(\left(1^2\right)^3\right)^1\right)^{\cdots}\right)^{2014}=1\]