Complex Numbers and Quadratic Equations — MATHS STD 11 Science — Question

Alternative Answer
$\sum\limits^{13}_{\text{n}=1}\big(\text{i}^{\text{n}}+\text{i}^{\text{n}+1}\big)=\sum\limits^{13}_{\text{n}=1}(1+\text{i})\text{i}^{\text{n}}$
$=(1+\text{i})(\text{i}+\text{i}^2+\text{i}^3+\text{i}^4+\text{i}^5+\text{i}^6+\text{i}^7+\text{i}^8+\text{i}^9+\text{i}^{10}+\text{i}^{11}+\text{i}^{12}+\text{i}^{13})$
$=(1+\text{i})\frac{\text{i}(\text{i}^{{13}}-1)}{\text{i}-1}=(1+\text{i})\frac{\text{i}(\text{i}-1)}{\text{i}-1}=(1+\text{i})\text{i}=\text{i}+\text{i}^2=\text{i}-1$