第 12 題小弟提供另一種作法。
第 12 題:
\(\displaystyle \sum_{k=1}^n (2k-1)^5\)
\(\displaystyle = \sum_{k=1}^n (2k)^5 + \sum_{k=1}^n \left(-5(2k)^4+10(2k^3)-10(2k)^2+5(2k)-1\right)\)
\(\displaystyle =32\sum_{k=1}^n k^5+O(n^5)\)
\(\displaystyle =32\left(\frac{1}{6}n^6+O(n^5)\right)+O(n^5)\)
\(\displaystyle =\frac{32}{6}n^6+O(n^5)\) (當 \(n\to\infty\) )
所以,\(\displaystyle \lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^n(2k-1)^5}{n^6}=\frac{16}{3}.\)
註:Big-O 定義請見
http://en.wikipedia.org/wiki/Big_O_notation