Question

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

Answer 1

Here, this series is in Geometric Progression (G.P).So, its value can be determined using the summation formula of G.P.

Formula used: ${\text{Sum of 'n' terms in a G}}{\text{.P = }}$ ${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$,
Where ,a: First Term , r : Common Ratio

Complete step-by-step answer:
Given: Series is 1+${i^2} + {i^4} + {i^6} + .......... + {i^{2n}}$
The Following series is in Geometric Progression , as Common ratio ‘r’ is the same for all terms.
Here, a is representing First term i.e,a = 1 and r is representing Common ratio i.e.,$r = \dfrac{{{i^2}}}{1} = \dfrac{{{i^4}}}{{{i^2}}} = {i^2}$
As, we know that sum of ‘n’ no. of terms of G.P =${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$
Putting the value of a=1 & r=${i^2}$, we have :
${S_n} = \dfrac{{1\left( {1 - {{\left( {{i^2}} \right)}^n}} \right)}}{{1 - {i^2}}} \\\ \\\$
$\Rightarrow {S_n} = \dfrac{{1 - {{\left( {{i^2}} \right)}^n}}}{{1 - \left( { - 1} \right)}} \\\ \Rightarrow {S_n} = \dfrac{{1 - {{\left( {{i^2}} \right)}^n}}}{2} \\\$
As, we know that value of ${i^2}$=-1.So, we will put its value in above equation.
$\Rightarrow {S_n} = \dfrac{{1 - {{\left( { - 1} \right)}^n}}}{2}$ …………………….(i)
Now, the value of ${S_n}$ will depend on the value of ‘n’.
Case -1 : When the value of ‘n’ is odd:
We will take the value of n =1, then we have :
${S_1} = \dfrac{{1 - {{\left( { - 1} \right)}^1}}}{2} \\\ \Rightarrow {S_1} = \dfrac{{1 + {1^{}}}}{2} = \dfrac{2}{2} = 1 \\\$
We will take the value of n =3, then we have:
${S_3} = \dfrac{{1 - {{\left( { - 1} \right)}^3}}}{2} = \dfrac{{1 - \left( { - 1} \right)}}{2} = \dfrac{2}{2} = 1$
Thus, we have observed for all odd values of ‘n’ , sum=1
${\text{i}}{\text{.e,}}$ ${S_n}$=1
Case 2: when the value of ‘n’ is even:
We will take the value of n =2, then we have:
${S_2} = \dfrac{{1 - {{\left( { - 1} \right)}^2}}}{2} = \dfrac{{1 - 1}}{2} = 0$
We will take the value of n =4, then we have:
${S_4} = \dfrac{{1 - {{\left( { - 1} \right)}^4}}}{2} = \dfrac{{1 - 1}}{2} = 0$
Thus, we have observed for all even values of ‘n’ , sum=0
${\text{i}}{\text{.e,}}$ ${S_n}$=0
Thus, the answer is 0 or 1, which will depend on whether the value of ‘n’ is even or odd.
So, unless the value of ‘n’ is given or specified, the exact answer can’t be determined.

So, the answer is Option:(d) cannot be determined.

Note: Geometric Progression (G.P) is a sequence in which each term is found by multiplying the previous term by a constant. Generally, G.P is like this: $a,ar,a{r^2},a{r^3},........,a{r^n}$ ,
Where, a=First term, r= common ratio