Question:
The value of $\left(i^{5}+i^{6}+i^{7}+i^{8}+i^{9}\right) /(1+i)$ is
(a) $\frac{1}{2}(1+i)$
(b) $\frac{1}{a}(1-i)$
(c) 1
(d) $\frac{1}{2}$
Solution:
(a) $\frac{1}{2}(1+i)$
$\frac{i^{5}+i^{6}+i^{7}+i^{8}+i^{9}}{1+i}$ $\left[\right.$ As, $\left.i^{5}=i, i^{6}=-1, i^{7}=-i, i^{8}=1, i^{9}=i\right]$
$=\frac{i-1-i+1+i}{1+i}$
$=\frac{i}{i+1}$
$=\frac{i}{i+1} \times \frac{i-1}{i-1}$
$=\frac{i(i-1)}{i^{2}-1}$
$=\frac{i^{2}-i}{-2}$
$=\frac{1}{2}(1+i)$
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