Question

$\sum_{k=0}^{440} i^k = x + iy \Rightarrow x^{100} + x^{99} y + x^{342} y^2 + x^{97} y^3 =$

• A. $0$
• B. $-4$
• C. $4$
• D. $1$

Hint:

Powers of $i$ repeat every four terms; use this to simplify large sums involving $i^k$.

Answer 1

The Correct Option is D

The powers of $i$ cycle every 4: $i^0 = 1$, $i^1 = i$, $i^2 = -1$, $i^3 = -i$, then repeat.
From $k=0$ to $440$, there are 441 terms.
Number of complete cycles = $\left\lfloor \frac{441}{4} \right\rfloor = 110$ cycles with 1 remainder.
So, total sum: $110(1 + i -1 - i) + i^0 = 110(0) + 1 = 1$
Thus, $x + iy = 1 \Rightarrow x = 1,\ y = 0$
Now plug into the expression: $x^{100} + x^{99} y + x^{342} y^2 + x^{97} y^3 = 1^{100} + 1^{99} \cdot 0 + 1^{342} \cdot 0 + 1^{97} \cdot 0 = 1$