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.