Sequence with imaginary numbers

Question

How do I solve this:

$$1 - i + i^2 - i^3 + i^4 + ... + i^{100} - i^{101}$$

I see that any 4 consecutive members of the sequence equal $0$.
If I extract $1$ and $-i^{101}$, I see there are 100 members, and since 100 is divisible by 4, these equal $0$.

What I'm left with is $$1-i^{101}=1-i,$$

but that's not the result. What did I do wrong?

Answer 1

Your sum is

$$\sum_{k=0}^{101}(-i)^k=\frac{1-(-i)^{102}}{1+i}=\frac{2}{1+i}=1-i$$

So your result is correct.

Answer 2

$$S=1 - i + i^2 - i^3 + i^4 + ... + i^{100} - i^{101}$$

$$Si=i - i^2 + i^3 - i^4 + i^5 + ... + i^{101} - i^{102}$$

Sum both:

$$S(i+1)=1-i^{102}$$ $$S=\frac{(1-i^{102})}{(i+1)}=\frac{(1-i^{102})(1-i)}{2}$$

Since $i^{100}=1$ then $i^{102}=-1$ then: $$S=\frac{2(1-i)}{2}=1-i$$

Answer 3

Pretty sure that you are right. Doing the same idea a different way,

$i^{4k}-i^{4k+1}+i^{4k+2}-i^{4k+3}=1-i+(-1)-(-i)$ pretty clearly equals zero, so all the terms except the last two go away. $i^{100}=1$ and $i^{101}=i$ so I'm pretty sure you're right.

Answer 4

Rewrite it on the geometric sum

$$\sum_{k=0}^{101}(-1)^ki^k=\frac{1-(-i)^{102}}{1+i}=\frac{2}{1+i}=1-i$$

Answer 5

It's the sum of the first $\;102\;$ elements of a geometric series that begins with $\;1\;$ and has $\;-i\;$ as constant ratio, thus its sum is

$$S:=1\cdot\frac{(-i)^{102}-1}{-i-1}=\frac{-2}{-1-i}=2\frac1{1+i}\frac{1-i}{1-i}=1-i$$