求数列{n(n+1)(n+2)}的前n项和sn
设数列{n(n+1)(n+2)(n+3)}的前n项和为Tn Tn=1×2×3×4+2×3×4×5+3×4×5×6+…+n(n+1)(n+2)(n+3)Tn= 1×2×3×4+2×3×4×5+…+(n-1)n(n+1)(n+2)+n(n+1)(n+2)(n+3)相减得0=1×2×3×4+4[2×3×4+3×4×5+n(n+1)(n+2)]-n(n...
一道高中数列题 数列{n(n+1)(n+2)(n+3)}的前n项和为 请写出具体过程谢谢...
a(n)=(n+3)!\/(n-1)!而不是a(n)=(n+3)! - (n-1)!显然, a(1)=4! 不等于 4!- 0!=4!-1 --- a(n)= n(n+1)(n+2)(n+3) = (1\/5)[n(n+1)(n+2)(n+3)(n+4) - (n-1)n(n+1)(n+2)(n+3) ]a(n-1)=(1\/5) [ (n-1)n(n+1)(n+2)(n+3)...
求数列{1\/n(n+1)(n+2)}的前n项和Sn 怎么计算
简单分析一下即可，详情如图所示
...的前n项和. 解:an=n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]\/3(裂项) 则...
=3n(n+1)于是 n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]\/3
求n(n+1)(n+2)(n+3)的前n项和
1*2*3*4=(1*2*3*4*5-0*1*2*3*4)\/5 2*3*4*5=(2*3*4*5*6-1*2*3*4*5)\/5 n(n+1)(n+2)(n+3)=[n(n+1)(n+2)(n+3)(n+4)-(n-1)n(n+1)(n+2)(n+3)]\/5 n(n+1)(n+2)(n+3)的前n项和 =(1*2*3*4*5-0*1*2*3*4)\/5+2*3*4*5*6-1*...
已知数列an=1\/n(n+1)(n+2),求数列的前n项和Sn 最好利用裂项法_百度知 ...
=1\/2{[(n+2)\/[n(n+1)(n+2)]-n\/[n(n+1)(n+2)]=1\/2{1\/[n(n+1)]-1\/[(n+1)(n+2)]} 所以Sn=1\/2*{1\/1*2-1\/2*3+1\/2*3-1\/3*4+……+1\/[n(n+1)]-1\/[(n+1)(n+2)]} =1\/2*{1\/1*2-1\/[(n+1)(n+2)]} =(n²+3n)\/(2n²+6n+4...
n(n+1)(n+2)数列求和
n(n+1)(n+2)=k(k+1)(k+2)(k+3)(k+4)\/ 先化简 =2N +N=3n乘n=2k+k 乘3k+9=7k+9 当N=1,3乘1=3，则7k+9=3所以成立 望采纳， Ps. 助人为乐乃人之本性，为大家解题是必要的，有不懂的要追问哦，对本人最大的支持和鼓励是 点赞哦，祝进步！
求数列{1\/n(n+1)(n+2)}的前n项和Sn
通项公式为An＝1\/n(n+1)(n+2)＝[1\/n(n+1)-1\/(n+1)(n+2)]\/2 所以Sn＝[1\/1*2－1\/2*3]\/2+ [1\/2*3－1\/3*4]\/2+[1\/3*4－1\/4*5]\/2+ ……+[1\/(n-1)*n－1\/n*(n+1)]\/2+ [1\/n(n+1)-1\/(n+1)(n+2)]\/2 ＝[1\/1*2-1\/(n+1)(n+2)]\/2 ＝1...
1.求通项公式为an=(2n+1)\/{n(n+1)(n+2)}的数列前n项和Sn 2.数列{an}...
=5\/4 - [4n+5]\/[2(n+1)(n+2)]a(n)=a(n-3),a(n+3)=a(n),a(3n-2)=a(1)=1,a(3n-1)=a(2)=3,a(3n)=a(3)=5.s(3n)=[a(1)+a(2)+a(3)]+...+[a(3n-2)+a(3n-1)+a(3n)]=[1+3+5]n=9n.s(3n-1)=s(3n)-a(3n)=9n-5.s(3n-2)=s(3n-1)-...
(1)求数列{(2n+1).2n}的前n项和 (2)求数列{(n+1)(n+2)\/1}的前n项
=n(n+1)[4n+8 - 3]\/3 =n(n+1)(4n+5)b(n)=1\/[(n+1)(n+2)] = 1\/(n+1) - 1\/(n+2),b(1)+b(2)+b(3)+...+b(n-1)+b(n)=1\/2-1\/3 + 1\/3-1\/4 + 1\/5-1\/4 + ... + 1\/n-1\/(n+1) + 1\/(n+1)-1\/(n+2)=1\/2 - 1\/(n+2)=(n+2-2)...
