Iotic iota #2

Algebra Level 2

∑ n = 0 1000 i n = ? \displaystyle\sum_{n=0}^{1000} i^n=? where i = − 1 i=\sqrt{-1} .

0 1 -1 None of the above 2 -2

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Parth Lohomi by Parth Lohomi , 20, India

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4 solutions

Chew-Seong Cheong
May 14, 2015

∑ n = 0 1000 i n = 1 + i − 1 − i + 1 + i − 1 − i + 1 + i − 1 − i . . . + 1 = + 1 \begin{aligned} \sum_{n=0}^{1000} {i^n} & = \color{#3D99F6}{1} \color{#D61F06}{+i}\color{#3D99F6}{-1} \color{#D61F06}{-i} \color{#3D99F6}{+1} \color{#D61F06}{+i}\color{#3D99F6}{-1} \color{#D61F06}{-i} \color{#3D99F6}{+1} \color{#D61F06}{+i}\color{#3D99F6}{-1} \color{#D61F06}{-i} ... \color{#3D99F6}{+1} \\ & = \boxed{\color{#3D99F6}{+1}} \end{aligned}

38 Helpful 2 Interesting 3 Brilliant 1 Confused

Moderator note:

Another way to solve this is to consider it as a geometric progression: 1 + i + i 2 + i 3 + … + i 1000 = i 1001 − 1 i − 1 = i 1001 m o d 4 − 1 i − 1 = i − 1 i − 1 = 1 \begin{aligned}1 + i + i^2 + i^3 + \ldots + i^{1000} &=& \frac{i^{1001} - 1}{i - 1} \\&=& \frac{i^{1001 \bmod \ 4} - 1}{i-1} \\ &=& \frac{i-1}{i-1} \\ &= &1 \end{aligned}

Adarsh Kumar
May 14, 2015

The sum of every four consecutive powers of i o t a iota is 0 0 .Now,that implies ∑ n = 1 1000 i n = 0 \sum_{n=1}^{1000} i^n=0 .Hence,answer is i 0 = 1 i^0=1 .1-minute problem!! Cheers.

21 Helpful 0 Interesting 1 Brilliant 2 Confused

Moderator note:

This works too!

well.. that was my approach... elementary but effective. :D

Rishabh Tripathi - 5 years, 11 months ago
Joe Potillor
Nov 14, 2016

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Mahmoud Meeda
Nov 3, 2020

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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