Question

Sn=1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2)求Sn

急！！！！

Answer 1

n*(n+1)*(n+2)=1/4*n*(n+1)*(n+2)[n+3-(n-1)]
Sn=1*2*3+2*3*4+3*4*5+...+n*(n+1)*(n+2)
=1/4{1*2*3*(4-0)+2*3*4*(5-1)+3*4*5*(6-2)...+n*(n+1)*(n+2)[n+3-(n-1)]}
如此裂项相消
原式= n*(n+1)*(n+2）*（n+3）/4

Answer 2

Sn’=1*2*3*4+2*3*4*5+...+(n-1)n(n+1)(n+2)+n(n+1)(n+2)(n+3) (1)
Sn'=1*2*3*4+...+(n-2)(n-1)n(n+1)+(n-1)n(n+1)(n+2)+n(n+1)(n+2)(n+3) (2)

(1)-(2)得：0=1*2*3*4+4*2*3*4+4*3*4*5+...+4(n-1)n(n+1)+4n(n+1)(n+2)-n(n+1)(n+2)(n+3)
0=4*{1*2*3+2*3*4+3*4*5+...+(n-1)n(n+1)+n(n+1)(n+2)}-n(n+1)(n+2)(n+3)
4*Sn=n(n+1)(n+2)(n+3)
所以Sn=n(n+1)(n+2)(n+3)/4

Answer 3

则an=(n+2)n=(n+1+1)(n+1-1)=(n+1)^2-1所以Sn=a1+a2+a3+a4+.....+anSn=(1+1)^2-1+(2+1)^2-1+(3+1)^2-1+...(n+1)^2-1= [2^2+ 3^2+4^2+...(n+1)^2]-n
=[2^2+3^2+4^2+...n^2+2n+1]-n=[1+2^2+3^2+.....n^2]+2n-n=[1+2^2+3^2+4^2+...n^2]+n
=[n(n+1)(2n+1)+6n]/6 在网上查得 1+
2^2+3^2+4^2+...n^2=[n(n+1)(2n+1)]/6不知道对不对，错了别怪我、、、

Answer 4

an=n(n+1)(n+2)=n^3+3n^2+2n
Sn=1^3+2^3+..n^3+3(1^2+2^2+...n^2)+2(1+2+...+n)
因为1^3+2^3+3^3+4^3+……+n^3=[n(n+1)/2]^2
1^2+2^2+3^2+4^4+……+n^2=
1+2+3+4+……+n=n(n+1)/2
Sn=[n(n+1)/2]^2+n(n+1)(2n+1)/2+n(n+1)

Answer 5

可以用积分算，你可以对x^3+x^4+……+x^（n+3），求和，然后把和函数求三次导，最后令x=1。