1×2+2×3+3×4+...+10×11 1×2+2×3+3×4+...+n×(n+1)
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/20 01:58:07
1×2+2×3+3×4+...+10×11 1×2+2×3+3×4+...+n×(n+1)
不要用什么很复杂的,我只是一个刚升初一的学生
不要用什么很复杂的,我只是一个刚升初一的学生
![1×2+2×3+3×4+...+10×11 1×2+2×3+3×4+...+n×(n+1)](/uploads/image/z/15741689-41-9.jpg?t=1%C3%972%2B2%C3%973%2B3%C3%974%2B...%2B10%C3%9711+1%C3%972%2B2%C3%973%2B3%C3%974%2B...%2Bn%C3%97%EF%BC%88n%2B1%EF%BC%89)
不会公式那就这么做
n(n+1)=1/3(n(n+1)(n+2)-(n-1)n(n+1))这叫消项n从取1开始
=1/3(1*2*3-0*1*2+2*3*4-1*2*3+.+10*11*12-9*10*11)
=1/3(10*11*12-0*1*2)
=440
同理
=1/3(1*2*3-0*1*2+2*3*4-1*2*3+.+n(n+1)(n+2)-(n-1)n(n+1))
=1/3(n(n+1)(n+2)-0*1*2)
=n(n+1)(n+2)/3
这种方法还可以延伸到1*2*3+2*3*4.
n(n+1)(n+2)=1/4(n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2))
再问: 嗯,谢谢你哦
再答: 这是初中解法
n(n+1)=1/3(n(n+1)(n+2)-(n-1)n(n+1))这叫消项n从取1开始
=1/3(1*2*3-0*1*2+2*3*4-1*2*3+.+10*11*12-9*10*11)
=1/3(10*11*12-0*1*2)
=440
同理
=1/3(1*2*3-0*1*2+2*3*4-1*2*3+.+n(n+1)(n+2)-(n-1)n(n+1))
=1/3(n(n+1)(n+2)-0*1*2)
=n(n+1)(n+2)/3
这种方法还可以延伸到1*2*3+2*3*4.
n(n+1)(n+2)=1/4(n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2))
再问: 嗯,谢谢你哦
再答: 这是初中解法
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
化简:1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)
(n+1)(n+2)/1 +(n+2)(n+3)/1 +(n+3)(n+4)/1
证明:1+2C(n,1)+4C(n,2)+...+2^nC(n,n)=3^n .(n∈N+)
一道数列求和题1/2n+3/4n+5/8n+...+(2n-1)/n*2^n
[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简
用数学归纳法证明:1×2×3+2×3×4+…+n×(n+1)×(n+2)=n(n+1)(n+2)(n+3)4(n∈N
若n为正整数,求1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)+.+1/
化简(n+1)(n+2)(n+3)
lim(1/n^2+4/n^2+7/n^2+…+3n-1/n^2)
求极限 lim(n->无穷)[(3n^2-2)/(3n^2+4)]^[n(n+1)]
求lim(n+1)(n+2)(n+3)/(n^4+n^2+1)