Question

The value of the sum $ \sum\limits_{n = 1}^{13} {\left( {{i^n} + {i^{n + 1}}} \right)} $ ,where $ i $ is the complex $ \sqrt { - 1} $ , is –
(A) $ i $
(B) $ - i $
(C) $ i + 1 $
(D) $ i - 1 $

Answer 1

Hint: We use all the value of iota $ \left( i \right) $ to solve such a type of question. The value of iota is $ \left( i \right) = \sqrt { - 1} $ , For reference also you should also remember $ {i^2} = - 1;{i^3} = - i\& {i^4} = 1 $ . Expand the summation sign and then use these values to simplify the equation.

Complete step-by-step answer:
In this question equation is given as
 $ \sum\limits_{n = 1}^{13} {\left( {{i^n} + {i^{n + 1}}} \right)} $ , $ i = \sqrt { - 1} $
We have to find the value of given equation
So,
 $ \sum\limits_{n = 1}^{13} {\left( {{i^n} + {i^{n + 1}}} \right)} $ . . . . . ( $ \sum\limits_{n = 1}^{13} {} $ means sum from $ n = 1 $ to $ n = 13 $ )
Since, summation is distributive over addition, on expanding the given equation, we get,
 $ \sum\limits_{n = 1}^{13} {\left( {{i^n} + {i^{n + 1}}} \right) = \left[ {\left( {\sum\limits_{n = 1}^{13} {{i^n}} } \right) + \sum\limits_{n = 1}^{13} {{i^{n + 1}}} } \right]} $
On applying the summation $ \left( {\sum {} } \right) $ rule on the above equation we get expression as,
 $ = \left[ {i + {i^2} + {i^3} + {i^4} + {i^5} + {i^6} + {i^7} + {i^8} + {i^9} + {i^{10}} + {i^{11}} + {i^{12}} + {i^{13}}} \right] + \left[ {{i^2} + {i^3} + {i^4} + {i^5} + {i^6} + {i^7} + {i^8} + {i^9} + {i^{10}} + {i^{11}} + {i^{12}} + {i^{13}} + {i^{14}}} \right] $
Let us call the above expression as S.
Now we know that $ i = \sqrt { - 1} $ .
According to the complex number property, “four consecutive terms of $ 'i' $ is zero”.
Proof of
 $ i + {i^2} + {i^3} + {i^4} = 0 $
We know $ i = \sqrt { - 1} $
Then
 $ {i^2} = - 1 $
By multiplying it by $ i $ we can write
 $ {i^3} = {i^2}.i = - i $
Then again, multiplying it by $ i $ , we can write
 $\Rightarrow {i^4} = {\left( {{i^2}} \right)^2} = {\left( { - 1} \right)^2} = 1 $
On adding the above four terms of $ 'i' $ we get
 $ i + {i^2} + {i^3} + {i^4} = i - 1 - i + 1 = 0 $
We can keep on doing the same thing to simplify all the powers of $ i $ in terms of $ \pm 1, \pm i $
Using the above expression, we can simplify S as
$\Rightarrow \left[ {i + {i^2} + {i^3} + {i^4} + {i^5} + {i^6} + {i^7} + {i^8} + {i^9} + {i^{10}} + {i^{11}} + {i^{12}} + {i^{13}}} \right] + \left[ {{i^2} + {i^3} + {i^4} + {i^5} + {i^6} + {i^7} + {i^8} + {i^9} + {i^{10}} + {i^{11}} + {i^{12}} + {i^{13}} + {i^{14}}} \right] $
\[ = \left[ {i - 1 - i + 1 + i - 1 - i + 1 + i - 1 - i + 1 + i} \right] + \left[ { - 1 - i + 1 + i - 1 - i + 1 + i - 1 - i + 1 + i - 1} \right]\]
By cancelling out the terms of opposite signs and simplifying the above equation, we get
 $ S = i - 1 $
Therefore, we get
 $ \sum\limits_{n = 1}^{13} {\left( {{i^n} + {i^{n + 1}}} \right)} = i - 1 $

Note: This question looks difficult but it is not. You just need to understand the values of different powers of $ i $ . Be careful while doing simplification. Do not make any mistake while simplifying the powers of $ i $ or cancelling out the terms.