问 题 | 推荐 收藏 |
提问:shdfjkhdsjkf 级别:二年级 来自:福建省 悬赏分:0
回答数:1 浏览数: |
|
结束时间:2014-05-05 11:15:33 问题补充 ┆ 评论 ┆ 举报 |
回 答 | 如果有了满意的回答请及时采纳,不要辜负了回答者 怎样采纳答案 |
回答:caiyunzuiyue 级别:幼儿园 2014-08-18 09:24:07 来自:河南省新乡市 |
答案:12253726。设<span lang="EN-US">1</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'">≤</span><span lang="EN-US">a</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'">≤</span><span lang="EN-US">99,<span lang="EN-US">1</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'">≤</span><span lang="EN-US">x</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'">≤</span><span lang="EN-US">a,对每一个固定的a和一个变化的x,在f(100,1)之前f(a,x)分别出现了a次,即f(a,x)共有a项,即f(1,x)有1项,f(2,x)有2项......f(99,x)有99项,即f(100,1)之前的项可简记为:f(1,x)(x=1),f(2,x)(<span lang="EN-US">1</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'">≤</span><span lang="EN-US">x</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'">≤</span><span lang="EN-US">a),f(3,x)(<span lang="EN-US">1</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'">≤</span><span lang="EN-US">x</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'">≤3</span><span lang="EN-US">)......f(99,x)(<span lang="EN-US">1</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'">≤</span><span lang="EN-US">x</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'">≤99</span><span lang="EN-US">),设f(100,1 )前共有t项,t=1+2+3+......+99=4950 。令f(1,1)=S<sub>1</sub>f(2,1)=S<sub>2,</sub>f(2,2)=S<sub>3,</sub>f(3,1)=S<sub>4,.</sub>.....所以f(100,1)=S<sub>t+1,</sub>=S<sub>4951,</sub>由题知:S<sub>1</sub>=1,S<sub>2</sub>=2,S<sub>3</sub>=4,S<sub>4</sub>=7,S<sub>5</sub>=11,S<sub>6</sub>=16,归纳出S<sub>n</sub>=1+[1+2+3+......+(n-2)+(n-1)]=1+[n*(n-1)]/2,S<sub>4951</sub>=1+(4950*4951) /2=12253726</span></span></span></span></span>
|
修改回答 ┆ 采纳答案 ┆ 评论 ┆ 举报 |
总回答数1,每页15条,当前第1页,共1页 |