Answer
423.9k+ views
Hint: First we’ll find different terms of the given expression, from there we’ll find the similarity, of the terms of the expression like some constant reoccurring terms.
After finding the expression we’ll find the value of the expression for different values of n, as n is not defined so it can be even or odd, so we make 2 cases and solve for both.
Complete step-by-step answer:
Given data: the expression $1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}}$
We know that, ${i^2} = - 1........(i)$
Squaring both the sides
$ \Rightarrow {i^4} = 1.......(ii)$
Multiplying equation(i) and (ii)
$ \Rightarrow {i^6} = - 1........(iii)$
Multiplying equation(i) and (iii)
$ \Rightarrow {i^8} = 1........(iv)$
on concluding from the above equation we can say that the expression terms are only 1 and -1 occurring alternatively
i.e. $1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}} = 1 + ( - 1) + 1 + ( - 1) + 1 + ( - 1) + ....{i^{2n}}$
now, we can write as ${i^{2n}} = {\left( {{i^2}} \right)^n}$
${i^{2n}} = {\left( { - 1} \right)^n}$
It is well known that ${\left( { - 1} \right)^a}$ is equal to 1 if a is even number and is equal to (-1) if a is odd
now if n is an even number in the expression, then the expression will end with 1
i.e. $1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}} = 1 + ( - 1) + 1 + ( - 1) + 1 + ( - 1) + ....1$
$\therefore 1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}} = 1$
if n is an odd number in the expression, then the expression will end with (-1)
i.e. $1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}} = 1 + ( - 1) + 1 + ( - 1) + 1 + ( - 1) + ....1 + ( - 1)$
$\therefore 1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}} = 0$
Therefore from the above equation, we conclude that the value of the given expression depends on the value on n and hence cannot be determined.
Option(D) is correct.
Note: An alternative method for this solution can be
We can see that the expression $1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}}$is a G.P. with a common ratio ${i^2}$
We know that the sum of first n terms of a G.P. with the first term as a and common ratio as r
${S_n} = a\dfrac{{\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}$
Therefore for the expression, $1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}} = 1\dfrac{{({{\left( {{i^2}} \right)}^{n + 1}} - 1)}}{{({i^2} - 1)}}$
Using ${i^2} = - 1$
$ = \dfrac{{({{\left( { - 1} \right)}^{n + 1}} - 1)}}{{( - 1 - 1)}}$
$ = \dfrac{{({{\left( { - 1} \right)}^{n + 1}} - 1)}}{{ - 2}}$
Now if n is even, (n+1) will be odd and ${\left( { - 1} \right)^{n + 1}} = - 1$
\[\therefore \dfrac{{({{\left( { - 1} \right)}^{n + 1}} - 1)}}{{ - 2}} = \dfrac{{( - 1 - 1)}}{{ - 2}}\]
\[ = 1\]
And if n is odd, (n+1) will be even and ${\left( { - 1} \right)^{n + 1}} = 1$
\[\therefore \dfrac{{({{\left( { - 1} \right)}^{n + 1}} - 1)}}{{ - 2}} = \dfrac{{(1 - 1)}}{{ - 2}}\]
\[ = 0\]
Hence, cannot be determined.
After finding the expression we’ll find the value of the expression for different values of n, as n is not defined so it can be even or odd, so we make 2 cases and solve for both.
Complete step-by-step answer:
Given data: the expression $1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}}$
We know that, ${i^2} = - 1........(i)$
Squaring both the sides
$ \Rightarrow {i^4} = 1.......(ii)$
Multiplying equation(i) and (ii)
$ \Rightarrow {i^6} = - 1........(iii)$
Multiplying equation(i) and (iii)
$ \Rightarrow {i^8} = 1........(iv)$
on concluding from the above equation we can say that the expression terms are only 1 and -1 occurring alternatively
i.e. $1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}} = 1 + ( - 1) + 1 + ( - 1) + 1 + ( - 1) + ....{i^{2n}}$
now, we can write as ${i^{2n}} = {\left( {{i^2}} \right)^n}$
${i^{2n}} = {\left( { - 1} \right)^n}$
It is well known that ${\left( { - 1} \right)^a}$ is equal to 1 if a is even number and is equal to (-1) if a is odd
now if n is an even number in the expression, then the expression will end with 1
i.e. $1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}} = 1 + ( - 1) + 1 + ( - 1) + 1 + ( - 1) + ....1$
$\therefore 1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}} = 1$
if n is an odd number in the expression, then the expression will end with (-1)
i.e. $1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}} = 1 + ( - 1) + 1 + ( - 1) + 1 + ( - 1) + ....1 + ( - 1)$
$\therefore 1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}} = 0$
Therefore from the above equation, we conclude that the value of the given expression depends on the value on n and hence cannot be determined.
Option(D) is correct.
Note: An alternative method for this solution can be
We can see that the expression $1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}}$is a G.P. with a common ratio ${i^2}$
We know that the sum of first n terms of a G.P. with the first term as a and common ratio as r
${S_n} = a\dfrac{{\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}$
Therefore for the expression, $1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}} = 1\dfrac{{({{\left( {{i^2}} \right)}^{n + 1}} - 1)}}{{({i^2} - 1)}}$
Using ${i^2} = - 1$
$ = \dfrac{{({{\left( { - 1} \right)}^{n + 1}} - 1)}}{{( - 1 - 1)}}$
$ = \dfrac{{({{\left( { - 1} \right)}^{n + 1}} - 1)}}{{ - 2}}$
Now if n is even, (n+1) will be odd and ${\left( { - 1} \right)^{n + 1}} = - 1$
\[\therefore \dfrac{{({{\left( { - 1} \right)}^{n + 1}} - 1)}}{{ - 2}} = \dfrac{{( - 1 - 1)}}{{ - 2}}\]
\[ = 1\]
And if n is odd, (n+1) will be even and ${\left( { - 1} \right)^{n + 1}} = 1$
\[\therefore \dfrac{{({{\left( { - 1} \right)}^{n + 1}} - 1)}}{{ - 2}} = \dfrac{{(1 - 1)}}{{ - 2}}\]
\[ = 0\]
Hence, cannot be determined.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)