Question

The value $1 + {i^2} + {i^4} + {i^6} + {i^8} + ..........{i^{2n}}$ is
A.Positive
B.Negative
C.Zero
D.Cannot be determined

Answer 1

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.