Question

If k, l, m, n are four consecutive integers, then ${{i}^{k}}+{{i}^{l}}+{{i}^{m}}+{{i}^{n}}$ is equal to
1. 1
2. 0
3. 2
4. 4

Answer 1

Hint: In this question, first we will represent k, l, m, n as consecutive numbers in terms of k. Then, we will use the properties of complex numbers to solve the problem. We will use the following properties to solve this problem.

Complete Step-by-Step solution:
$\begin{align}
  & \Rightarrow i=\sqrt{-1} \\
 & \Rightarrow {{i}^{2}}=-1 \\
 & \Rightarrow {{i}^{3}}=({{i}^{2}}\times i)=(-1\times i) \\
 & \Rightarrow {{i}^{3}}=-i \\
\end{align}$
We know that, since k, l, m, n are four consecutive integers it can be represented as,
$\Rightarrow l=k+1.......(i)$
$\Rightarrow m=k+2........(ii)$
$\Rightarrow n=k+3.......(iii)$
We have to find the value of ${{i}^{k}}+{{i}^{l}}+{{i}^{m}}+{{i}^{n}}$.
Substituting the values of k, l, m and n obtained from equations (i), (ii) and (iii) in the above expression we get,
$\Rightarrow {{i}^{k}}+{{i}^{l}}+{{i}^{m}}+{{i}^{n}}={{i}^{k}}+{{i}^{k+1}}+{{i}^{k+2}}+{{i}^{k+3}}...........(iv)$
Equation (iv) can be further simplified by taking ${{i}^{k}}$ as common. Then equation (iv) changes to,
$\Rightarrow {{i}^{k}}+{{i}^{l}}+{{i}^{m}}+{{i}^{n}}={{i}^{k}}(1+{{i}^{1}}+{{i}^{2}}+{{i}^{3}})...........(v)$
From the properties of complex numbers, we know that,
$\begin{align}
  & \Rightarrow i=\sqrt{-1} \\
 & \Rightarrow {{i}^{2}}=-1 \\
 & \Rightarrow {{i}^{3}}=({{i}^{2}}\times i)=(-1\times i) \\
 & \Rightarrow {{i}^{3}}=-i \\
\end{align}$
Therefore, comparing with these properties $i$,${{i}^{2}}$ and ${{i}^{3}}$ are substituted in equation (v).
Then equation (v) becomes,
$\Rightarrow {{i}^{k}}+{{i}^{l}}+{{i}^{m}}+{{i}^{n}}={{i}^{k}}(1+i-1-i)...........(vi)$
On simplifying equation (vi) we get,
$\begin{align}
  & \Rightarrow {{i}^{k}}+{{i}^{l}}+{{i}^{m}}+{{i}^{n}}={{i}^{k}}(0) \\
 & \therefore {{i}^{k}}+{{i}^{l}}+{{i}^{m}}+{{i}^{n}}=0........(vii) \\
\end{align}$
Thus, the correct answer is option 2.

Note: The first step involved in solving this question is writing k, l, m and n as consecutive integers in terms of k. The student should know how to represent consecutive integers. Here, l, m and n are represented as l = k + 1, m = k + 2 and n = k + 3. If this step becomes correct, the solution can be easily found out using the properties of complex numbers as mentioned in the solution.