Question

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

Answer 1

1*2+2*3+3*4+....+n(n+1)
=1/3(1*2*3-0*1*2+2*3*4-1*2*3+...+n(n+1)(n+2)-(n-1)n(n+1))
=1/3n(n+1)(n+2)
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+...+n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2))
=1/4n(n+1)(n+2)(n+3)

追问

为啥

追答

自己看啊,很详细的呀!

Answer 2

1*2+2*3+3*4+....+n(n+1)
=1/3(1*2*3-0*1*2+2*3*4-1*2*3+...+n(n+1)(n+2)-(n-1)n(n+1))
=1/3n(n+1)(n+2)
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+...+n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2))
=1/4n(n+1)(n+2)(n+3)
用数学归纳法证明一下！~