Question

The value of $ 1 + i^2 + i^4 + i^6 + -------- + i^{2n} $ is:
a). Positive
b). Negative
c). 0
d). Cannot be determined

Answer 1

Hint: The series given above in iota (“i”) is in G.P. so we are going to use the formula of summation of G.P. for n terms and also use the values of iota with different powers followed by the general algebra of simplification.

Complete step-by-step solution -
The series given above is in G.P. in which the first term is 1 and common ratio is $ i^2$.
$1 + i^2 + i^4 + i^6 + -------- + i^{2n}$
From the above series, the common ratio is calculated by dividing any term by its preceding term. So, let us take any term say $i^2$ dividing by its preceding term which is 1 we get the common ratio as $i^2$.
Now, formula for the summation of G.P. is:
${{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r}$
In the above expression, n represents the number of terms, “a” is the first term of a G.P. and “r” is the common ratio of a G.P.
There are n + 1 terms in the given sequence in iota. Substituting the value of a, r and n in the summation expression, we get:
$\begin{align}
  & {{S}_{n+1}}=\dfrac{1\left( 1-{{\left( {{i}^{2}} \right)}^{n+1}} \right)}{1-{{i}^{2}}} \\
 & \Rightarrow {{S}_{n+1}}=\dfrac{1-{{\left( -1 \right)}^{n+1}}}{1+1} \\
 & \Rightarrow {{S}_{n+1}}=\dfrac{1-{{\left( -1 \right)}^{n+1}}}{2} \\
\end{align}$
In the above steps, we have taken $i^2 = -1$. This is due to complex number iota property.
When n is odd then n + 1 is even and the value of $ (-1)^{n + 1} $ is +1 so the above expression value is:
${{S}_{n+1}}=0$
When n is even then n + 1 is odd and the value of $ (-1)^{n + 1} $ is -1 so the value of summation is:
$\begin{align}
  & {{S}_{n+1}}=\dfrac{1+1}{2} \\
 & \Rightarrow {{S}_{n+1}}=1 \\
\end{align}$
Hence, the value of summation of the sequence depends on the value of n so we cannot determine the value of summation unless n is specified.
Hence, the correct option is (d).

Note: Instead of taking the whole sequence as G.P., you can just take this sequence $ i^2 + i^4 + i^6 + -------- + i^{2n} $ as G.P. then find the summation of this G.P. and add 1 to it. There is a difference between the number of terms in the sequence of given problem and the note part. There are (n+1) terms in the solution part while there are n number of terms in sequence which we are assuming in the note part.