Complex Numbers
Complex Number in Iota Form
Complex Numbers_PYQ
Grade 11
Question:
The value of the sum $\displaystyle\sum_{n=1}^{13} (i^n + i^{n+1})$, where $i = \sqrt{-1}$, equals
Step-by-Step Solution
Key Concept: Factoring $(i^n + i^{n+1}) = i^n(1+i)$ converts the sum into a simple geometric series whose period-4 cancellation is easy to exploit.
**Step 1: Factor out**
$\displaystyle\sum_{n=1}^{13}(i^n + i^{n+1}) = (1+i)\sum_{n=1}^{13} i^n$.
**Step 2: Sum the geometric series**
Powers of $i$ cycle with period 4, each complete cycle summing to $i - 1 - i + 1 = 0$. For $n = 1$ to $12$: 3 complete cycles $\Rightarrow 0$. For $n = 13$: $i^{13} = i$. So $\sum_{n=1}^{13} i^n = i$.
**Step 3: Final answer**
$(1+i) \cdot i = i + i^2 = i - 1$.
Correct Answer: 2