王朝知道

分享

 

01.Sn=1*2*3+2*3*4+......+n(n+1)(n+2)
an=n(n+1)(n+2) =((n+3)-(n-1))/4*n(n+1)(n+2) =[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]/4 加起来，可以狂消Sn=[n(n+1)(n+2)(n+3)-0]/4 =n(n+1)(n+2)(n+3)/4...查看完整版>>Sn=1*2*3+2*3*4+......+n(n+1)(n+2)02.已知：an=n(n+1)(n+2) 求：Sn
n(n+1)(n+2) =3!*C(n+2,3)Sn=3!*[ C(3,3)+C(4,3)+C(5,3)+……+C(n+2,3)] =3!*C(n+3,4) =3!(n+3)!/4!(n-1)!=(n+3)(n+2)(n+1)n/4...查看完整版>>已知：an=n(n+1)(n+2) 求：Sn03.1*2*3+2*3*4+3*4*5+......+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4怎样推导出来的呢
1×2×3+2×3×4+3×4×5+......+n(n+1)(n+2)=1/4【1×2×3×4-0×1×2×3】+1/4【2×3×4×5-1×2×3×4】+1/4【3×4×5×6-2×3×4×5】+......+1/4【n(n+1)(n+2)（n+3)-(n-1)n(n+1)(n+2)】=1/4n(n+1)(n+2)(n+3)...查看完整版>>1*2*3+2*3*4+3*4*5+......+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4怎样推导出来的呢04.1*2+2*3+3*4+......+n(n+1)=n(n+1)(n+2)/3 是怎样推导出来的呢？
n*(n+1)=n*n+n 所以：1×2+2×3+3×4+4×5+┉┉ =（1*1+2*2+3*3+……）+（1+2+3+……） 前面用前n项的平方和公式，后面有前n项求和公式就可以了。进一步推导就可得出你要的结论...查看完整版>>1*2+2*3+3*4+......+n(n+1)=n(n+1)(n+2)/3 是怎样推导出来的呢？05.编写程序，求s=1×2×3 +2×3×4+……+n×(n+1)×(n+2)的值。n由文本框输入
Private Sub Command1_Click() Dim n As Integer, i As Integer, s As Long n = Val(Text1.Text) If n < 1 Then Label1.Caption = \"输入错误\" Else s = 0 For i = 1 To n Step 1 s...查看完整版>>编写程序，求s=1×2×3 +2×3×4+……+n×(n+1)×(n+2)的值。n由文本框输入06.10.用vb编写程序，求s=1×2×3 +2×3×4+……+n×(n+1)×(n+2)的值。n由文本框输入
Private Sub Command1_Click() Dim i As Integer, s As Long s = 0 For i = 1 To Int(Text1.Text) s = s + (i * (i + 1) * (i + 2)) Next i print sEnd Sub...查看完整版>>10.用vb编写程序，求s=1×2×3 +2×3×4+……+n×(n+1)×(n+2)的值。n由文本框输入07.求解1/1×2×3+1/2×3×4+……+1/n(n+1)(n+2)
1/n(n+1)(n+2)=1/n(n+2)-1/(n+1)(n+2)=(1/2)[1/n-1/(n+2)]-[1/(n+1)-1/(n+2)]所以原式＝(1/2)[1-1/3+1/2-1/4+1/3-1/5......+1/(n+1)-1/(n+2)]+[1/2-1/3+1/3-1/4+.........+1/(n+1)-1/(n+2)]＝(1/2)[1+1/2-1/(n+1)-1/...查看完整版>>求解1/1×2×3+1/2×3×4+……+1/n(n+1)(n+2)08.1/1*2*3+1/2*3*4+1/3*4*5+...+1/n*(n+1)*(n+2)=?
=(1-1/(n+2))/3...查看完整版>>1/1*2*3+1/2*3*4+1/3*4*5+...+1/n*(n+1)*(n+2)=?09.1/(n+1)+1/(n+2)+......+1/(n+n)
象数学问题...查看完整版>>1/(n+1)+1/(n+2)+......+1/(n+n)10.1*2*3+4*5*6+......n*(n+1)*(n+2)=?
#include <stdio.h>int main(){ int i, n, sum=0; printf("Enter a number"); scanf("%d", &n); for(i = 1; i <= n; i += 3) sum += i * (i+1) * (i+2); printf("%d\n", sum); return 0;}...查看完整版>>1*2*3+4*5*6+......n*(n+1)*(n+2)=?

 

 

免责声明：本文为网络用户发布，其观点仅代表作者个人观点，与本站无关，本站仅提供信息存储服务。文中陈述内容未经本站证实，其真实性、完整性、及时性本站不作任何保证或承诺，请读者仅作参考，并请自行核实相关内容。