Iota Terminator

Algebra Level 3

$\Large \sum_{k=1}^{3^{2n^3+8n+50} - 1} i^k$

Let $n$ denote a positive integer, and for $i = \sqrt{-1}$ , evaluate the summation above for all $n$ .

i

-1 0 1

-i

1 solution

Shivamani Patil
Jun 29, 2015

${ 3 }^{ 2n^{ 3 }+8n+50 }-1={ 3 }^{ 2(n^{ 3 }+4n+25) }-1={ 3 }^{ 2y }-1\quad$

By Euler's theorem we know that ${ 3 }^{ 2y }-1\equiv 0\quad$ mod $4$ .

$\Rightarrow { 3 }^{ 2n^{ 3 }+8n+50 }-1=4z$

It is known that sum of $4$ consecutive powers of $i$ results in $0$ .

Therefore our answer is Aryabhata's invention or $0$ .

Nice problem, but anyone can guess the solution because we have 3 attemptions.

It could be $1$ , $-1$ , $i$ , $-i$ or $0$ , you can't enter $i$ or $-i$ as answer so. :D

Вук Радовић - 5 years, 11 months ago

But probability that your answer is correct at first attempt is $\frac { 1 }{ 3 }$ :).

shivamani patil - 5 years, 11 months ago

Good point. I've added an MCQ format to this question.

Brilliant Mathematics Staff - 5 years, 11 months ago

@Brilliant Mathematics – Thank you it is better.

shivamani patil - 5 years, 11 months ago

You don't really need Euler's Theorem here. It can be done using basic modular arithmetic as follows:

$3^{2n^3+8n+50}-1\equiv (-1)^{2(n^3+4n+25)}-1\equiv 1^{n^3+4n+25}-1\pmod{4}$

Now, we note that $\forall~n\in\Bbb{Z^+}$ , we have $n^3+4n+25\in\Bbb{Z^+}$ and hence, we have,

$3^{2n^3+8n+50}-1\equiv 1-1\equiv 0\pmod{4}$

using the fact that $1^x=1~\forall~x\in\Bbb{Z^+}$ .

Prasun Biswas - 5 years, 11 months ago

That's right.But that time Euler came in my mind so is Euler's theorem.

shivamani patil - 5 years, 11 months ago