Question

Let i = √-1. If $\frac{(-1 + i\sqrt{3})^{21}}{(1 - i)^{24}} + \frac{(1 + i\sqrt{3})^{21}}{(1 + i)^{24}} = k$, and $n = \lfloor |k| \rfloor$. Then $\sum_{j=0}^{n+5} (j + 5)^2 - \sum_{j=0}^{n+5} (j + 5)$ is equal to __________.

Hint:

Always look for Euler's form $re^{i\theta}$ or Cube roots of unity ($\omega$) when dealing with high powers of complex numbers.

Answer 1

Correct Answer: 310

Step 1: $-1+i\sqrt{3} = 2\omega$ (where $\omega = e^{i2\pi/3}$). $(2\omega)^{21} = 2^{21}(\omega^3)^7 = 2^{21}$.
Step 2: $(1-i)^2 = -2i \Rightarrow (1-i)^{24} = (-2i)^{12} = 2^{12} \cdot 1 = 2^{12}$.
Step 3: First term $= 2^{21}/2^{12} = 2^9 = 512$.
Step 4: $1+i\sqrt{3} = -2\omega^2$. $(-2\omega^2)^{21} = -2^{21}(\omega^3)^{14} = -2^{21}$.
Step 5: $(1+i)^2 = 2i \Rightarrow (1+i)^{24} = (2i)^{12} = 2^{12}$.
Step 6: Second term $= -2^{21}/2^{12} = -512$. Thus $k = 512 - 512 = 0$.
Step 7: $n = 0$. Summation $\sum_{j=0}^{5} (j+5)^2 - \sum_{j=0}^{5} (j+5) = \sum_{j=0}^{5} (j+5)(j+4)$. Actually, checking the powers: if $k=0$, the expression simplifies significantly. (Note: If $k$ was a large number, we would use $\sum n^2$ formulas).