Question

Consider $i=\sqrt{-1}$ , then the value of the sum $\sum\limits_{n=1}^{16}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}$ will be equal to:
(a) i
(b) i – 1
(c) -i
(d) 0

Answer 1

Hint: Find the conditions possible on n. As given n is an integer the cases possible are n is odd and n is even. So, check the values in both the cases.

Complete step-by-step solution -
Definition of i, can be written as:
The solution of the equation: ${{x}^{2}}+1=0$ is i. i is an imaginary number. Any number which has an imaginary number in its representation is called a complex number.
Definition of a complex equation, can be written as: An equation containing complex numbers in it, is called a complex equation.
It is possible to have a real root for complex equations.
Example: $\left( 1+i \right)x+\left( 1+i \right)=0,x= -1$ is the root of the equation.
Expression given in the question:
$\sum\limits_{n=1}^{16}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}$
Case 1: Let the value of n be 1
By substituting this into equation, we get:
$\begin{aligned} & =i+{{i}^{1+1}}=i+{{i}^{2}} \\\ & =i-1 \\\ \end{aligned}$
Case 2: Let the value of n be 2
By substituting this into equation, we get:
$\begin{aligned} & ={{i}^{2}}+{{i}^{2+1}}={{i}^{2}}+{{i}^{3}} \\\ & =-1-i \\\ \end{aligned}$
Case 3: Let the value of n be 3
By substituting this into equation, we get:
$\begin{aligned} & ={{i}^{3}}+{{i}^{3+1}}={{i}^{3}}+{{i}^{4}} \\\ & =-1+i \\\ \end{aligned}$
Case 4: Let the value of n be 4
By substituting this into equation, we get:
$\begin{aligned} & ={{i}^{4}}+{{i}^{5}} \\\ & =1+i \\\ \end{aligned}$
Case 5: Let the value of n be 5
By substituting this into equation, we get:
$\begin{aligned} & ={{i}^{5}}+{{i}^{6}} \\\ & =i-1 \\\ \end{aligned}$
Case 6: Let the value of n be 6
By substituting this into equation, we get:
$\begin{aligned} & ={{i}^{6}}+{{i}^{7}} \\\ & =-1-i \\\ \end{aligned}$
Case 7: Let the value of n be 7
By substituting this into equation, we get:
$\begin{aligned} & ={{i}^{7}}+{{i}^{8}} \\\ & =-1+i \\\ \end{aligned}$
Case 8: Let the value of n be 8
By substituting this into equation, we get:
$\begin{aligned} & ={{i}^{8}}+{{i}^{9}} \\\ & =1+i \\\ \end{aligned}$
Every case is repeating itself after 4 times. So, we can write
$\sum\limits_{n=1}^{16}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}=4\sum\limits_{n=1}^{4}{{{i}^{n}}+{{i}^{n+1}}}$
Case 1 + Case 2 + Case 3 + Case 4
By substituting we get: = $i-1-1-i-i+1+1+i=0$
$\sum\limits_{n=1}^{16}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}=4\left( 0 \right)=0$
Hence 0 is the value of the expression
Option (d) is correct.

Note: The idea of seeing the cyclicity of all cases after every 4 cases is important. Do it carefully. Along with the definition of a complex number students must know about the property of iota. Here we can use an alternate method by putting the value of n from 1 to 16. we will get a series, then by the property of iota we get the suitable answer.