Called Complex For A Reason - ii

Algebra Level 2

$\large i^1 + i^2 + i^3 + \cdots + i^{98} + i^{99} + i^{100} = \, ?$

Clarification : $i = \sqrt{-1}$ .

i^{(101\times50)}

50i

100i

3 solutions

Rishabh Jain
Feb 23, 2016

$\color{#D61F06}{\mathcal{METHOD}~ 1}$ This represents a geometric progression sum , $i^1 + i^2 + i^3 +\cdots + i^{98} + i^{99} + i^{100}$ $=\dfrac{i(i^{100}-1)}{i-1}=0$ (Since $i^{100}=i^4=1$ ) $\color{#D61F06}{\mathcal{METHOD}~ 2}$ Since $i^2=-1, i^3=-i, i^4=1$ $\implies i+i^2+i^3+i^4=0$ Hence starting from the first term every four consecutive terms add to zero. Hence $25\times 4=100$ will also add up to 0.

Rohit Udaiwal
Feb 23, 2016

We use here the formula for finite G.P.,in which $a=i,;r=i;n=100$ $\displaystyle \sum_{n=1}^{100} i^n=\dfrac{i(1-i^{100})}{1-i} =\dfrac{i\cdot (1-1)}{1-i}=0$

I'm not getting the solution

Narayan Palsikar - 5 years, 3 months ago

we see that in the method 2 first 4 consecutive terms when added gives 0.....i^1+i^2+i^3+i^4=> i+(-1)+(-i)+1=> 0........Since 100 is a multiple of 4...it produces the answer 0 at last...

Keerthana Balan - 5 years, 3 months ago

A Former Brilliant Member
Feb 28, 2016

Learn how to math and 0

Called Complex For A Reason - ii

3 solutions

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