Question

Evaluate the value of $1+{{i}^{2}}+{{i}^{4}}+{{i}^{6}}+.....+{{i}^{2n}}$
(a) positive
(b) negative
(c) zero
(d) cannot be determined

Answer 1

Hint: First, before proceeding for this, we must know the following values of powers of i as${{i}^{1}}=i,{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$.Then, if we suppose that n is odd, then values of n can 1, 3, 5, .. and so on, we get the value as (-1). Then, we suppose that n is even, then values of n can be 2, 4, 6, .. and so on, we get the value as 1. Then, the result of the summation of series $1+{{i}^{2}}+{{i}^{4}}+{{i}^{6}}+.....+{{i}^{2n}}$ can be 0 or 1 unless n is specified exact answer cannot be determined.

Complete step by step answer:
In this question, we are supposed to find the value of series as $1+{{i}^{2}}+{{i}^{4}}+{{i}^{6}}+.....+{{i}^{2n}}$.
So, before proceeding for this, we must know the following values of powers of i as:
$\begin{align}
  & {{i}^{1}}=i \\
 & {{i}^{2}}=-1 \\
 & {{i}^{3}}=-i \\
 & {{i}^{4}}=1 \\
\end{align}$
Then, we also know that this cycle repeats after the power of 4 and we get the same values of ${{i}^{5}}$as i, ${{i}^{6}}$as (-1), ${{i}^{7}}$as (-i) and ${{i}^{8}}$as1.
Now, we need to find the value of the series till ${{i}^{2n}}$ where it is not given that n is even or odd.
Now, if we suppose that n is odd, then values of n can 1, 3, 5, .. and so on.
Then, let us calculate the value of ${{i}^{2n}}$when n is odd as:
${{i}^{2n}}=-1$
So, we get the value of ${{i}^{2n}}$ as (-1) when n is odd.
Similarly, if we suppose that n is even, then values of n can 2, 4, 6, .. and so on.
Then, let us calculate the value of ${{i}^{2n}}$when n is even as:
${{i}^{2n}}=1$
So, we get the value of ${{i}^{2n}}$ as 1 when n is even.
Now, if n is odd then the last term will be (-1) and hence the summation becomes as:
$\begin{align}
  & 1+{{i}^{2}}+{{i}^{4}}+{{i}^{6}}+.....+{{i}^{2n}} \\
 & \Rightarrow 1+\left( -1 \right)+1+\left( -1 \right)+......+\left( -1 \right) \\
 & \Rightarrow 0 \\
\end{align}$
Then, if n is even then the last term will be 1 and hence the summation becomes as:
$\begin{align}
  & 1+{{i}^{2}}+{{i}^{4}}+{{i}^{6}}+.....+{{i}^{2n}} \\
 & \Rightarrow 1+\left( -1 \right)+1+\left( -1 \right)+......+1 \\
 & \Rightarrow 1 \\
\end{align}$
So, we can clearly see that if n is even, the result of the summation is 0.
Also, we can clearly see that if n is even, the result of the summation is 1.
So, the result of the summation of series $1+{{i}^{2}}+{{i}^{4}}+{{i}^{6}}+.....+{{i}^{2n}}$ can be 0 or 1 unless n is specified exact answer cannot be determined.

So, the correct answer is “Option D”.

Note: Now, to solve these types of questions we need to know some of the basic things about $i$ as it is used to represent an imaginary part of a complex number in the form as $a+ib$. Moreover , we must know the value of $i$as $\sqrt{-1}$.