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Question: The value of \( 1 + i^2 + i^4 + i^6 + -------- + i^{2n} \) is: a). Positive b). Negative c). 0...

The value of 1+i2+i4+i6+−−−−−−−−+i2n1 + i^2 + i^4 + i^6 + -------- + i^{2n} is:
a). Positive
b). Negative
c). 0
d). Cannot be determined

Explanation

Solution

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 i2 i^2.
1+i2+i4+i6+−−−−−−−−+i2n1 + 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 i2i^2 dividing by its preceding term which is 1 we get the common ratio as i2i^2.
Now, formula for the summation of G.P. is:
Sn=a(1−rn)1−r{{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:
Sn+1=1(1−(i2)n+1)1−i2 ⇒Sn+1=1−(−1)n+11+1 ⇒Sn+1=1−(−1)n+12 \begin{aligned} & {{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{aligned}
In the above steps, we have taken i2=−1i^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(-1)^{n + 1} is +1 so the above expression value is:
Sn+1=0{{S}_{n+1}}=0
When n is even then n + 1 is odd and the value of (−1)n+1(-1)^{n + 1} is -1 so the value of summation is:
Sn+1=1+12 ⇒Sn+1=1 \begin{aligned} & {{S}_{n+1}}=\dfrac{1+1}{2} \\\ & \Rightarrow {{S}_{n+1}}=1 \\\ \end{aligned}
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 i2+i4+i6+−−−−−−−−+i2ni^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.