n!! 1×2＋2×3×4……＋n（n＋1）（n＋2）=

1×2＋2×3×4……＋n（n＋1）（n＋2）=  

1×2＋2×3×4……＋n（n＋1）（n＋2）=

1×2×3＝1/4[1×2×3×4-0×1×2×3]
2×3×4=1/4[2×3×4×5-1×2×3×4]
3×4×5=1/4[3×4×5×6-2×3×4×5]
.........
n（n+1）(n+2)＝1/4[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]
所以1×2×3+2×3×4+3×4×5+...+n(n+1)(n+2)
＝1/4[1×2×3×4-0×1×2×3+2×3×4×5-1×2×3×4+3×4×5×6-2×3×4×5+.....+n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]
=1/4n(n+1)(n+2)(n+3)

1*2+2*3+……+n(n+1)(N+2)=

Sn=1×2+2×3+3×4+……+n(n+1)=1²+1+2²+2+3²+3+……+n²+n
=(1+2+3+……+n)+(1²+2²+3²+……+n²)
=n(n+1)/2+(1²+2²+3²+……+n²)
关键求1²+2²+3²+……+n²如下 1²+2²+3²+……+n²=n(n+1)(2n+1)/6
证:（利用恒等式(n+1)³=n³+3n²+3n+1）：
(n+1)³-n³=3n²+3n+1,
n³-(n-1)³=3(n-1)²+3(n-1)+1
..............................
3³-2³=3×2²+3×2+1
2³-1³=3×1²+3×1+1.
把这n个等式两端分别相加，得：
(n+1)³-1=3(1²+2²+3²+……+n²)+3(1+2+3+...+n)+n,
由于1+2+3+...+n=n(n+1)/2,
代人上式得：
n³+3n²+3n=3(1²+2²+3²+……+n²)+3(n+1)n/2+n
整理后得：
1²+2²+3²+……+n²=n(n+1)(2n+1)/6
所以Sn=n(n+1)/2+(1²+2²+3²+……+n²)=n(n+1)/2+n(n+1)(2n+1)/6

=n(n²+3n+2)/3

由下列等式，你有什么猜想？1+2+3+4+5+…+n=1/2*n(n+1);1*2+2*3+3*4+…+n*(n+1)=1/3*n*(n+1)*(n+2)

猜想
1×2×3+2×3×4+3×4×5+……n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4
证明:
n(n+1)(n+2)=n^3+3n^2+2n
1×2×3+2×3×4+3×4×5+……n(n+1)(n+2)
=(1^3+2^3+……+n^3)+3*(1^2+2^2+……+n^2)+2*(1+2+……+n)
1^3+2^3+……+n^3=[n(n+1)/2]^2
1^2+2^2+……+n^2=n(n+1)(2n+1)/6
1+2+……+n=n(n+1)/2
所以原式=[n(n+1)/2]^2+3n(n+1)(2n+1)/6+2n(n+1)/2
=n^2(n+1)^2/4+n(n+1)(2n+1)/2+n(n+1)
=[n(n+1)/4]*[n(n+1)+2(2n+1)+4]
=[n(n+1)/4]*(n^2+5n+6)
=n(n+1)(n+2)(n+3)/4

用数学归纳法证明: 1*2+2*3+3*4+.+n(n+1)=1/3n(n+1)(n+2)

当n=1时，1*2=1*(1+1)*(1+2)/3, 该等式成立

现在假设n=k时，1*2+2*3+...+k(k+1)=k(k+1)(k+2)/3 成立 k为自然数

则当n=k+1时，1*2+2*3+...+k(k+1)+(k+1)(k+2)
=k(k+1)(k+2)/3+(k+1)(k+2)
=(k/3+1)(k+1)(k+2)
=(k+1)(k+1+1)(k+1+2)/3

即，当n=k+1时也成立

所以该等式对所有n都成立

化简：3/(1!+2!+3!)+4/(2!+3!+4!)+.+（n+2)/[n!+(n+1)!+(n+2)!]

先看一般项。
(n+2)/[n!+(n+1)!+(n+2)!]
=(n+2)/[n!*(1+(n+1)+(n+1)(n+2))]
=(n+2)/[n!*(n+2)^2]
=1/[n!(n+2)]
=(n+1)/[n!(n+1)(n+2)]
=(n+1)/(n+2)!
=(n+2-1)/(n+2)!
=1/(n+1)!-1/(n+2)!
所以
3/(1!+2!+3!)+4/(2!+3!+4!)+...+(n+2)/[n!+(n+1)!+(n+2)!]
=(1/2!-1/3!)+(1/3!-1/4!)+...+(1/(n+1)!-1/(n+2)!)
=1/2-1/(n+2)!
即原式=1/2-1/(n+2)!

证明1÷(1×2×3)+1÷(2×3×4)+.+1÷(N(N+1)(n+2)<1÷4?

1/k(k+1)(k+2)=(1/2)*(1/k(k+1)-1/(k+1)(k+2))（可以把右边通分出来验证一下）
所以：原式
=1/(1*2*3)+1/(2*3*4)+...+1/n(n+1)(n+2)
=(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)*(1/2-1/(n+1)(n+2))
=1/4-1/2(n+1)(n+2)<1/4

1·n+2·（n-1）+3·（n-2）+……n·1=1/6·n（n+1）（n+2）

证明：
（1）当n=1时，左边=1*1=1，右边=（1/6）*1*2*3=1
左边=右边，等式成立。
（2）假设当n=k时，等式成立。
即 1*k+2*(k-1)+……+(k-1)*2+k*1=(1/6)*k*(k+1)*(k+2)
当n=k+1时
左边=1*(k+1)+2*k+……+(k-1)*3+k*2+（k+1）*1
=[1*k+1*1]+[2*(k-1)+2]+……+[(k-1)*2+k-1]+[k*1+k]+(k+1)
(把每一项分成两项）
=[1*k+2*(k-1)+……+(k-1)*2+k*1]+[1+2+……+(k-1)+k+(k+1)]
(前一项是把n=k时成立的结果带进去，后一项是等差数列)
=(1/6)*k*(k+1)*(k+2)+(1/2)(k+1)(k+2)
=(1/6)(k+1)(k+2)(k+3)
即当n=k+1时等式也成立。
综上，原等式恒成立。

如何证明1*2x3+2x3x4+.+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4

1*2x3+2x3x4+......+n(n+1)(n+2)=（2²-1）×2+（3²-1）×3+……+【（n+1）²-1】（n+1）
=2³+3³+……+（n+1）³-2-3-……-（n+1）
=1³+2³+3³+……+（n+1）³-（1+2+3+……+n+1）
=（1+2+3+……+n+1）²-（1+2+3+……+n+1）
=（1+2+3+……+n+1）（2+3+……n+1）
=（1+n+1）（n+1）/2 *（2+n+1）n/2
=（n+2）（n+1）（n+3）n/4