Question

What is the value of the sum
$\sum\limits_{n = 2}^{11} {\left( {{i^n} + {i^{n + 1}}} \right)} {\text{ where i = }}\sqrt { - 1}$ ?
A. i
B. 2i
C. -2i
D. 1+i

Answer 1

Hint: In this question, we will use the results of power of ‘i’. We will use the results given as:
${{\text{i}}^{4n}} = 1,{\text{ }}{{\text{i}}^{4n + 1}} = i,{\text{ }}{{\text{i}}^{4n + 2}} = - 1{\text{ and }}{{\text{i}}^{4n + 3}} = - i.$ We will first expand the summation and then use these results to get the answer.

Complete step-by-step solution -

In the given question, we have to find the value of summation given below:
S=$\sum\limits_{n = 2}^{11} {\left( {{i^n} + {i^{n + 1}}} \right)}$
First we will expand this summation as follow:
S= ${{\text{i}}^2} + {{\text{i}}^3} + {{\text{i}}^3} + {{\text{i}}^4} + {{\text{i}}^4} + {{\text{i}}^5} + {{\text{i}}^5}{\text{ + }}{{\text{i}}^6}{\text{ + }}{{\text{i}}^6}{\text{ + }}{{\text{i}}^7}{\text{ + }}{{\text{i}}^7}{\text{ + }}{{\text{i}}^8}{\text{ + }}{{\text{i}}^8}{\text{ + }}{{\text{i}}^9}{\text{ + }}{{\text{i}}^9}{\text{ + }}{{\text{i}}^{10}}{\text{ + }}{{\text{i}}^{10}}{\text{ + }}{{\text{i}}^{11}}{\text{ + }}{{\text{i}}^{11}}{\text{ + }}{{\text{i}}^{12}} \\\ = {{\text{i}}^2} + 2({{\text{i}}^3} + {{\text{i}}^4}{\text{ + }}{{\text{i}}^5}{\text{ + }}{{\text{i}}^6}{\text{ + }}{{\text{i}}^7}{\text{ + }}{{\text{i}}^8}{\text{ + }}{{\text{i}}^9}{\text{ + }}{{\text{i}}^{10}}{\text{ + }}{{\text{i}}^{11}}{\text{) + }}{{\text{i}}^{12}} \\\$
Now we that:
${{\text{i}}^{4n}} = 1 \\\ {\text{ }}{{\text{i}}^{4n + 1}} = i \\\ {\text{ }}{{\text{i}}^{4n + 2}} = - 1{\text{ }} \\\ {\text{and }}{{\text{i}}^{4n + 3}} = - i. \\\$
Using these results, we get:
S = -1 +2(-i + 1 + i + (-1) + (-i) + 1 + i + (-1) +(-i)) + 1= -2i.
Therefore, option C is correct.

Note: In the problem involving summation on complex numbers, you must remember the standard exponent results given below:
${{\text{i}}^{4n}} = 1 \\\ {\text{ }}{{\text{i}}^{4n + 1}} = i \\\ {\text{ }}{{\text{i}}^{4n + 2}} = - 1{\text{ }} \\\ {\text{and }}{{\text{i}}^{4n + 3}} = - i. \\\$
It will be helpful in solving problems related to summation of complex numbers having different indices(powers).