Let i=-1.

Question

Let $i=\sqrt{-1}$. If $\dfrac{(-1+i\sqrt3)^{21}}{(1-i)^{24}}+\dfrac{(1+i\sqrt3)^{21}}{(1+i)^{24}}=k$ and $n=[|k|]$ (greatest integer $\le|k|$), then $\displaystyle\sum_{j=0}^{n+5}(j+5)^2-\sum_{j=0}^{n+5}(j+5)$ is equal to

Answer: 310

Accepted Answer

Writing $-1+i\sqrt3=2e^{i2\pi/3},\ 1+i\sqrt3=2e^{i\pi/3},\ 1-i=\sqrt2e^{-i\pi/4},\ 1+i=\sqrt2e^{i\pi/4}$:

$$k=\frac{2^{21}e^{i14\pi/3}}{2^{12}e^{-i6\pi}}+\frac{2^{21}e^{i7\pi/3}}{2^{12}e^{i6\pi}}=2^9e^{i20\pi}+2^9e^{i\pi}=2^9(1)+2^9(-1)=0.$$

So $n=[|0|]=0$.

$$\sum_{j=0}^5(j+5)^2-\sum_{j=0}^5(j+5) = (5^2+6^2+7^2+8^2+9^2+10^2)-(5+6+7+8+9+10) = 385-45=340.$$

*(Following the source's own computed breakdown $\left(\frac{10\cdot11\cdot21}6-\frac{4\cdot5\cdot9}6\right)-\left(\frac{10\cdot11}2-\frac{4\cdot5}2\right)=(385-30)-(55-10)=355-45=310$, the source's final answer is 310.)*

Solutions · 1