Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

If ${a_n} = {i^{{{(n + 1)}^2}}}$ , where $i = \sqrt { - 1} $ and n=1,2,3 ... n, then the value of ${a_1} + {a_3} + {a_5} + \,...\, + {a_{25}}$ is
A) $13$
B) $13 - i$
C) $13 + i$
D) $12$
D) $12 - i$

Answer
VerifiedVerified
509.4k+ views
Hint: So, they have asked to find the sum of series of ${a_1} + {a_3} + {a_5} + \,...\, + {a_{25}}$ and they have given ${a_n} = {i^{{{(n + 1)}^2}}}$. Now, to find the solution for the given problem, first you need to see how many terms are present in it. Then put their term values in ${a_n} = {i^{{{(n + 1)}^2}}}$ this equation and remember that they also given one more thing and that it $i = \sqrt { - 1} $ and after remembering this you just need to put value and calculate it to get correct answer.

Complete step by step answer:
First of all they given,
$ \Rightarrow {a_n} = {i^{{{(n + 1)}^2}}}$
Now, we have to look at the given series. Given series is,
$ \Rightarrow {a_1} + {a_3} + {a_5} + \,...\, + {a_{25}}$
If you look at the series you will figure out that only $13$ terms are present in it. Now, count this series as $13$ odd number as we can see. So, we can put values of it in ${a_n} = {i^{{{(n + 1)}^2}}}$ this equation and we will get following terms,
$ \Rightarrow {a_1} + {a_3} + {a_5} + \,...\, + {a_{25}} = {i^{{{(1 + 1)}^2}}} + {i^{{{(3 + 1)}^2}}} + {i^{{{(5 + 1)}^2}}} + \,...\, + {i^{{{(25 + 1)}^2}}}$
From further simplification,
$ \Rightarrow {a_1} + {a_3} + {a_5} + \,...\, + {a_{25}} = {i^{(4)}} + {i^{(16)}} + {i^{(36)}} + \,...\, + {i^{(676)}}$
Now, as we can have given that $i = \sqrt { - 1} $ so we can clearly say that ${i^4} = 1$
So as there are $13$ terms so we can rewrite above equation as bellow,
$ \Rightarrow {a_1} + {a_3} + {a_5} + \,...\, + {a_{25}} = 1 + 1 + 1 + \,...\, + 1$
So there are $13$ terms so the sum of the given series is $13$ .
Therefore, There are 13 terms in the series. Hence, option (A) is the correct option.

Note:
In this question they have given $i = \sqrt { - 1} $ so this is nothing but it is a complex number. It contains imaginary numbers and $i$ is called “iota”. Basically the complex number is the combination of a real number and an imaginary number. Hence, a complex number is a simple representation of the addition of two numbers, i.e., real number and an imaginary number.