(b) We know that, sum of any four consecutive powers of $i$ is zero $\therefore\ \ \ \ \sum_{n = 1}^{13}{\left( i^{n} + i^{n + 1} \right) = \left( i + i^{2} + .... + i^{13} \right) + (i^{2} + i^{3} + ... + i^{14})}$ $= i^{13} + i^{14}$ $= i - 1$
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Ph.D. in Physics with 19+ years of teaching experience.