Question

The value of $\frac{i^5+i^6+i^7+i^8+i^9}{1+i}$ is

• A. $1$
• B. $\frac{1}{2}(1-i)$
• C. $\frac{1}{2}$
• D. $\frac{1}{2}(1+i)$

Answer 1

The correct option is D $\frac{1}{2}(1+i)$

$\because \frac{i^5+i^6+i^7+i^8+i^9}{1+i}$

$=\frac{i^{(4+1)}+i^{(4+2)}+i^{(4+3)}+i^{(4\times 2+0)}+i^{(4\times 2+1))}}{1+i}$

$=\frac{(i^1+i^2+i^3+i^0+i^1)}{(1+i)}$

$\{\because i^{(4\times n+k)}=i^k\}$

$=\frac{(i-1-i+1+i)}{(1+i)}$

$=\frac{i}{1+i}$

$=\frac{i}{1+i}\times \frac{1-i}{1-i}$

$=\frac{i-i^2}{1^2-i^2}$

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

$=\frac{1}{2}(i+1)$

$\therefore \frac{i^5+i^6+i^7+i^8+i^9}{1+i}=\frac{1}{2}(i+1)$
Hence, option (A) is correct.