回復 1# ibvtys 的帖子
第12題
考慮 \(\displaystyle (\frac{1}{2}-\omega^{k})(\frac{1}{2}-\omega^{n-k})=(\frac{1}{2})^{2}-(\omega^{k}+\omega^{n-k})\cdot\frac{1}{2}+1=\frac{5}{4}-(\frac{\omega^{k}}{2}+\frac{1}{2\omega^{k}})=\frac{5}{4}-\frac{\omega^{2k}+1}{2\omega^{k}}\)
因為 \(\displaystyle 1+\frac{1}{2}+\frac{1}{2^{2}}+......+\frac{1}{2^{n-1}}=(\frac{1}{2}-\omega)(\frac{1}{2}-\omega^{2})......(\frac{1}{2}-\omega^{n-1})\)
\(\displaystyle =(\frac{1}{2}-\omega^{n-1})(\frac{1}{2}-\omega^{n-2})...(\frac{1}{2}-\omega)\)
將此式連乘兩次得(等式後面前後對調上下相乘)
所求 \(\displaystyle A_{n}=(1+\frac{1}{2}+\frac{1}{2^{2}}+......+\frac{1}{2^{n-1}})^{2}\)
即 \(\displaystyle \lim\limits_{n\rightarrow\infty}A_{n}=(\frac{1}{1-\frac{1}{2}})^{2}=4\)
