Question

If ${(1+i)}^{2n}+{(1-i)}^{2n}=-2^{n+1}$ where, $i=\sqrt{-1}$ for all those $n$, which are

• A. even
• B. odd
• C. multiple of $3$
• D. None of these

Answer 1

The correct option is A even

In $(1+i{)}^{2n}+(1-i{)}^{2n}$

$={\left\{\right(1+i{)}^2\}}^n+{\left\{\right(1-i{)}^2\}}^n$

$=(1+i^2+2i{)}^n+(1+i^2-2i{)}^n$

$=(1-1+2i{)}^n+(1-1-2i{)}^n$

$=2^n{i}^2+i^2(-2{)}^n$

$=i^2(2^2+(-2{)}^n)$

When $n=2$

$=i^2(2^2+2^2)=-1\cdot 2^{2+1}$

Hence, ${}^{\prime }{n}^{\prime }$ must be even