引用:
原帖由 armopen 於 2009-5-23 10:50 AM 發表 
請問第 6 題除了硬算之外有比較好的作法嗎？ 謝謝！
求 1^2 * C(10,1) * (1/6) * (1/6)^9 + 2^2 * C(10,2) * (1/6)^2 * (1/6)^8 + ... + 10^2 C(10,10) * (1/6)^10 = ?
拋磚引玉一下
\(\displaystyle (x+y)^n=\sum_{k=0}^{n}C_{k}^{n}x^{k}y^{n-k}\)
考慮
\(\displaystyle \sum_{k=1}^{n}kC_{k}^{n}x^{k}y^{n-k}\)
\(\displaystyle =\sum_{k=1}^{n}nC_{k-1}^{n-1}x^{k}y^{n-k} \)
\(\displaystyle =nx\sum_{k=0}^{n-1}C_{k}^{n-1}x^{k}y^{n-1-k} \)
\(\displaystyle =nx(x+y)^{n-1} \)
所以
\(\displaystyle 1*C_{1}^{10}(\frac{1}{6})(\frac{5}{6})^{9}+2*C_{2}^{10}(\frac{1}{6})^{2}(\frac{5}{6})^{8}+...+10*C_{10}^{10}(\frac{1}{6})^{10} \)
\(\displaystyle =10*(\frac{1}{6})(\frac{1}{6}+\frac{5}{6})^{9}=\frac{10}{6} \)
再考慮
\(\displaystyle \sum_{k=2}^{n}k(k-1)C_{k}^{n}x^{k}y^{n-k} \)
\(\displaystyle =\sum_{k=2}^{n}n(n-1)C_{k-2}^{n-2}x^{k}y^{n-k} \)
\(\displaystyle =n(n-1)x^2\sum_{k=0}^{n-2}C_{k}^{n-2}x^{k}y^{n-2-k} \)
\(\displaystyle =nx^2(x+y)^{n-2} \)
所以
\(\displaystyle (1^2-1)*C_{1}^{10}(\frac{1}{6})(\frac{5}{6})^{9}+(2^2-2)*C_{2}^{10}(\frac{1}{6})^{2}(\frac{5}{6})^{8}+...+(10^2-10)*C_{10}^{10}(\frac{1}{6})^{10} \)
\(\displaystyle =10*9*(\frac{1}{6})^{2}(\frac{1}{6}+\frac{5}{6})^{8}=\frac{15}{6} \)
由上面兩式得到求值式為 \(\displaystyle \frac{15}{6}+\frac{10}{6}=\frac{25}{6} \)
初次嘗試使用jsMath發文，如果有錯誤或是關於Sigma的表示應該用別的方式，請瑋岳老師指導一下
還是可以直接從我的電腦上傳圖檔??
