好像可以用Σ算，但是我不會
更新1:
其實不用Σ算也可以啦
高中矩陣的運算問題
回答 (2)
2012-04-19 5:43 am
2012-04-20 3:04 am
a(i,j) = i^2+J^2
40n^2 = ΣΣa(i,j) = ΣΣ(i^2+j^2) = nΣi^2+nΣj^2 = 2n(1^2+2^2+...+n^2)
=2n*n(n+1)(2n+1)/6 = n^2(n+1)(2n+1)/3
故 40=(n+1)(2n+1)/3. 解之.
40n^2 = ΣΣa(i,j) = ΣΣ(i^2+j^2) = nΣi^2+nΣj^2 = 2n(1^2+2^2+...+n^2)
=2n*n(n+1)(2n+1)/6 = n^2(n+1)(2n+1)/3
故 40=(n+1)(2n+1)/3. 解之.
收錄日期: 2021-05-04 01:50:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120418000016KK06575