Question

数学题求助，在线等答案！！

用数学归纳法证明
1*2*3+2*3*4+……+n(n+1)(n+2)=(1/4)n(n+1)(n+2)(n+3)
数学归纳法，注意

Answer 1

数学归纳法就很简单了- -
首先，1*2*3=（1/4)*1*2*3*4成立，
1*2*3+2*3*4=(1/4)*2*3*4*5也成立，
假设当n=k时1*2*3+2*3*4+……+k(k+1)(k+2)=(1/4)k(k+1)(k+2)(k+3)成立
则当n=k+1时1*2*3+2*3*4+……+(k+1)(k+2)(k+3)=(1/4)k(k+1)(k+2)(k+3)+(k+1)(k+2)(k+3)=((1/4)k+1)(k+1)(k+2)(k+3)=(1/4)(k+1)(k+2)(k+3)(k+4)
也成立，以此类推，对于任意正整数n，1*2*3+2*3*4+……+n(n+1)(n+2)=(1/4)n(n+1)(n+2)(n+3) 都成立~

Answer 2

设Sn=1*2*3+2*3*4+……+n(n+1)(n+2),
那么
4Sn=1*2*3*4+2*3*4*4+……+4n(n+1)(n+2)
=2*3*4*(1+4)+3*4*5*4+……+4n(n+1)(n+2)
=2*3*4*5+3*4*5*4+……+4n(n+1)(n+2)
=3*4*5*(2+4)+4*5*6*4+……+4n(n+1)(n+2)
=3*4*5*6+5*6*7*4+……+4n(n+1)(n+2)
……
=n(n+1)(n+2)(n+3)
那么Sn=1/4n(n+1)(n+2)(n+3)
我是老师 谢谢采纳

Answer 3

Sn=1/4n(n+1)(n+2)(n+3)