1. The problem asks to find the value of the sum $$1 + i + i^2 + i^3 + i^4 + \cdots + i^{16}$$ where $i$ is the imaginary unit with the property $i^2 = -1$. 2. Important property: Powers of $i$ cycle every 4 steps: $$i^0 = 1, \quad i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad \text{and then it repeats}.$$ 3. We can group the terms in blocks of 4: $$ (i^0 + i^1 + i^2 + i^3) + (i^4 + i^5 + i^6 + i^7) + \cdots + (i^{12} + i^{13} + i^{14} + i^{15}) + i^{16} $$ 4. Each group of 4 terms sums to zero: $$1 + i + (-1) + (-i) = 0$$ 5. There are 4 such groups from $i^0$ to $i^{15}$, so their sum is: $$4 \times 0 = 0$$ 6. Now add the last term $i^{16}$: Since $i^{16} = (i^4)^4 = 1^4 = 1$ 7. Therefore, the total sum is: $$0 + 1 = 1$$ Final answer: 1