pure maths evaluate急
回答 (2)
2008-01-14 4:35 am
✔ 最佳答案
1sq-3sq+5sq-7sq+...+(4n+1)sq=1+(5-3)(5+3)+(9-7)(9+7)+...+((4n+1)-(4n-1))((4n+1)+(4n-1))
=1+2(8+16+...+8n)
=1+16(1+2+...+n)
=1+16n(n+1)/2
=1+8n(n+1).
2008-01-14 4:40 am
1sq-3sq+5sq-7sq+......-(4n-1)sq+(4n+1)sq
= summation(i=0 to i=n)(4i+1)sq - summation(i=1 to i=n)(4i-1)sq
= 1+ summation(i=1 to i=n)((4i+1)sq-(4i-1)sq)
=1 + summation(i=1 to i=n)(16i)
=1 + 16n(n+1)/2
=1+ 8n(n+1)
=8nsq+8n+1
= summation(i=0 to i=n)(4i+1)sq - summation(i=1 to i=n)(4i-1)sq
= 1+ summation(i=1 to i=n)((4i+1)sq-(4i-1)sq)
=1 + summation(i=1 to i=n)(16i)
=1 + 16n(n+1)/2
=1+ 8n(n+1)
=8nsq+8n+1
參考: my self
收錄日期: 2021-04-13 14:57:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080113000051KK03974