1x2+2x3+3x4+4x5+.+n(n+1)等于多少?急救! 请写下过程,谢谢

如题所述

第1个回答  2019-04-11
1x2+2x3+3x4+…+n(n+1) =1^2+1+2^2+2+3^2+3+…+n^2+n =(1^2+2^2+3^2+…+n^2)+(1+2+3+…+n) =1/6*n(n+1)(2n+1)+1/2*n(n+1) =1/6*n(n+1)(2n+1+3)(提取公因式) =1/3*n(n+1)(n+2)

1x2+2x3+3x4+4x5+.+n(n+1)等于多少?急救! 请写下过程,谢谢
1x2+2x3+3x4+…+n(n+1) =1^2+1+2^2+2+3^2+3+…+n^2+n =(1^2+2^2+3^2+…+n^2)+(1+2+3+…+n) =1\/6*n(n+1)(2n+1)+1\/2*n(n+1) =1\/6*n(n+1)(2n+1+3)(提取公因式) =1\/3*n(n+1)(n+2)

简算1x2+2x3+3x4+4x5+...+n(n+1)
1x2+2x3+3x4+4x5+...+n(n+1)=(1^2+2^2+……n^2)+(1+2+3+……n)=n(n+1)(2n+1)\/6+(1+n)xn\/2 =n(n+1)(n+2)\/3

1x2+2x3+3x4+...+ n的计算方法
求1x2+2x3+3x4+……+n(n+1)注意到:(n+1)^3-n^3=3n^2+3n+1 则可得:n(n+1)=[(n+1)^3-n^3]\/3-1\/3 那么有:1×2=(2^3-1^3)\/3-1\/3 2×3=(3^3-2^3)\/3-1\/3 ……累加可得:所求算式 =[(n+1)^3-1^3]\/3-n\/3 =(n^3+3n^2+3n-n)\/3 =n(n+1)...

1×2+2x3+3x4+4x5如何简算?
根据公式可计算:n(n+1)=n^2+n 1×2+2x3+3x4+4x5 =(1^2+2^2+...5^2)+(1+2+...+5)=(1+4+9+16+25)+(1+2+...+5)=55+15 =70 混合运算:如果一级运算和二级运算,同时有,先算二级运算。如果一级,二级,三级运算(即乘方、开方和对数运算)同时有,先算三级运算...

1x2+2x3+3x4+4x5+...+n(n+1)+n(n+2)=?(过程—)
3n*(n+1) = n(n+1)(n+2) - (n-1)n(n+1)所以3S = (n+1)(n+2)(n+3) - 0 S = (n+1)(n+2)(n+3) \/3

1乘2+2乘3+3乘4+···+99乘100=?
= [ 1X2X3 -1X2X3 +2X3X4 -2X3X4 +3X4X5 -3X4X5 +4X5X6 ] \/3 = 4X5X6 \/3 规律你看出来了吗?这个数列的公式就是 通项 a= n(n+1)前n项数列和 S= n(n+1)(n+2)\/3 这样一来,一直加到 99X100,就是 1X2 +2X3 +3X4 +4X5 +……+99X100 = 99X100X101 \/3 = ...

1X2X3+2X3X4+3X4X5+...+N(N+1)(N+2) 奥数题 要简便方法不要那个用微...
回答:是不是这个公式 (n-1)n(n+1)=n^3-n {n^3}求和公式:Sn=[n(n+1)\/2]^2 {n}求和公式:Sn=n(n+1)\/2

1x2+2x3+3x4+4x5+···n(n+1)=___(n为自然数)
n(n+1)=n^2+nS(n)=(1+2^2+3^2+...+n^2)+(1+2+3+...+n)=n(n+1)(2n+1)\/6+n(n+1)\/2 =n(n+1)(2n+1+3)\/6 =n(n+1)(n+2)\/3

1X2+2X3+3X4+4X5...+2011X2013=
1X2+2X3+3X4+4X5...+n(n+1)=(1²+2²+3²+---+n²)+(1+2+3+---+n)=n(n+1)(2n+1)\/6+n(n+1)\/2 =(1\/6)n(n+1)(2n+1+3)=n(n+1)(n+2)\/3 把n=2012代入得:1X2+2X3+3X4+4X5...+2012X2013 =2012*2013*2014\/3=2012*671*2014=27...

为什么1*2+2*3+3*4+4*5+……+n*(n+1)=n*(n+1)*(n+2)\/3 谁能给我个证明...
解析:因为1×2=1\/3×1×2×3,1×2+2×3=1\/3×2×3×4,1×2+2×3+3×4=1\/3×3×4×5,1×2+2×3+3×4+4×5=1\/3×4×5×6,结论:1×2+2×3+3×4+…+n(n+1)= n(n+1)(n+2)这种题的规律很难发现【解析】这个主要利用两个公式1+2+3+...+...

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