Question

If $S\left(n\right)=i^n+i^{-n}$, where $i=\sqrt{-1}$ and $n$ is a positive integer, then the total number of distinct values of $S\left(n\right)$ is:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

$i=i$

$i^2=-1$

$i^3=-i$

$i^4=1$

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

$i^{-n}$ can be written as,

$\frac{1}{i^n}={\left(\frac{1}{i}\right)}^n={\left(\frac{1\times i}{i\times i}\right)}^n={\left(\frac{i}{-1}\right)}^n$

$={(-i)}^n$

$\therefore S_n=i^n+{(-i)}^n$

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

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

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

if $n=$odd ; then $S_n=0$

if $n=$even , we have 2 cases,

(i) $n$ is multiple of $4\Rightarrow S_n=1(1+1)=2$

(ii) $n$ is NOT a mulple of $4\Rightarrow S_n=-1(1+1)=-2$

Thus three distinct values of $S_n=\{-2,0,2\}$