The value of the series 1 + i^2 + i^4 + i^6 + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given series is $S = 1 + i^2 + i^4 + i^6 + ..... + i^{2n}$.Since $i^2 = -1$,$i^4 = 1$,$i^6 = -1$,and so on,the series becomes $S = 1 - 1 + 1 -

Vedclass · Accepted Answer

Cannot be determined

Vedclass · Answer

(D) The given series is $S = 1 + i^2 + i^4 + i^6 + ..... + i^{2n}$.Since $i^2 = -1$,$i^4 = 1$,$i^6 = -1$,and so on,the series becomes $S = 1 - 1 + 1 - 1 + ..... + (-1)^n$.This is a finite geometric series with $n+1$ terms,where the first term $a = 1$ and the common ratio $r = i^2 = -1$.The sum of a geometric series is $S = \frac{a(1 - r^{n+1})}{1 - r}$.Substituting the values,$S = \frac{1(1 - (-1)^{n+1})}{1 - (-1)} = \frac{1 - (-1)^{n+1}}{2}$.If $n$ is even,$n+1$ is odd,so $S = \frac{1 - (-1)}{2} = \frac{2}{2} = 1$.If $n$ is odd,$n+1$ is even,so $S = \frac{1 - 1}{2} = 0$.Since the value depends on the parity of $n$,it cannot be determined without knowing $n$.

The value of the series $1 + i^2 + i^4 + i^6 + ..... + i^{2n}$ is:

Similar Questions

The inequality $a + ib > c + id$ is meaningful only when:

If $x, y \in R$ and $(x + iy)(3 + 2i) = 1 + i$,then $(x, y)$ is

If one root of the quadratic equation $ix^2 - 2(i + 1)x + (2 - i) = 0$ is $2 - i$,then the other root is

Express $(-\sqrt{3}+\sqrt{-2})(2 \sqrt{3}-i)$ in the form of $a+ib$.

Express the given complex number in the form $a+ib$: $\left(\frac{1}{5}+i \frac{2}{5}\right)-\left(4+i \frac{5}{2}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module