Question

$- 1$will be real, if $- 3$ =

• A. $1 + i^{2} + i^{4} + i^{6} + ..... + i^{2n}$
• B. $i^{2} + i^{4} + i^{6} + ......$
• C. $(2n + 1)$
• D. None of these [Where $i$ is an integer]

Answer 1

Answer: (C)

$(2n + 1)$

Answer 2

$\mathbf{=}\frac{\mathbf{i(1 -}\mathbf{i}^{\mathbf{1000}}\mathbf{)}}{\mathbf{1 - i}}\mathbf{=}$= $\frac{i(1 - (i^{4})^{250})}{1 - i}$

Now, since it is real, therefore Im $= \frac{i(1 - 1)}{1 - i} = 0$

⇒ $x^{2} - 6x + 10 = 0$= 0 ⇒ $x^{3} - 3x^{2} - 8x + 15$, $= x(x^{2} - 6x + 10) + 3(x^{2} - 6x + 10) - 15 = x(0) + 3(0) - 15 = - 15$

Where $(1 + i)^{2n} = (1 - i)^{2n}$, 1, 2, 3, ......

Trick : Check for (1), if $\Rightarrow \left( \frac{1 + i}{1 - i} \right)^{2n} = 1$ the given number is absolutely real but (3) also satisfies this condition and in (1) and (3), (3) is most general value of $\Rightarrow (i)^{2n} = 1$.