Demonstrating that $\sum_{n=0}^{50}i^n=i$ I got this summation and need to demonstrate why the result is $i$. Any ideas?
$$\sum_{n=0}^{50}i^n=i$$
 A: First, this is simply a brain teaser than a linear algebra problem. Note that $i$ imaginary number has the following traits: $i^2=-1$, $i^3=-i$ and $i^4=1$. Given this, Note that summation of all even indices of $n=4*j$ where $j$ some non-negative integers and $n=4*j+2$ are going to cancel each other out up to $n=50$. This means even terms will give you $0$ contribution to the sum. Now consider odd-numbered indices, then you know that there are $25$ odd terms that cancel each other out because they are alternating $-i$ and $i$ except one term is not canceled. Odd indexed terms then contribute $i$ to the sum. Hence, you end up with a total sum of $i$.
A: When dealing with powers of $i$, the following are important:
$i^1 = i$,
$i^2 = -1$,
$i^3 = -i$,
$i^4 = 1$,
So, $i$ raised to any power greater than $4$ can be written in terms of these 4 terms. For example, $i^5$ can be written as $i^4 \cdot i = i$. Now, when you try to find the the values of every four terms in the summation (starting from $n = 1$) , you will realise that the sum of every four terms is $0$. So,
$\sum_{n=1}^{4}i^n=0$, $\sum_{n=5}^{8}i^n=0$, $\sum_{n=9}^{12}i^n=0$, and so on..
Therefore, $\sum_{n=1}^{48}i^n=0$
Now that we've dealt with $48$ terms, we are left with the $0^{th}$, $49^{th}$ and $50^{th}$, which are $1, -1$ and $i$ respectively. 
$1-1+i = i$
Thus,
$\sum_{n=0}^{50}i^n= 1 +\sum_{n=1}^{48}i^n -1 +i = i$
A: $$i^0+i^1+i^2+i^3=1+i-1-i=0$$
$$i^4+i^5+i^6+i^7=1+i-1-i=0$$
$$\vdots$$
$$i^{44}+i^{45}+i^{46}+i^{47}=1+i-1-i=0$$
$$i^{48}+i^{49}+i^{50}=1+i-1=i$$
A: Since $i^2=-1,i^3=-i,i^4=1$, every 4 consecative terms cancle out. 
So$\sum\limits_{n=0}^{50}i^n=\sum\limits_{n=48}^{50}i^n=1+i+i^2=i$ 
A: Geometric series:
$S_{50}=\dfrac{1-i^{51}}{1-i}=$
$(1/2))(1-i^{51})(1+i) =$
$(1/2)(1+i)^2=i.$
Used:
$i^{51}=i^{48}i^3=1(-i)$
