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

Question

说详细点

Answer 1

证明：1×2+2×3+3×4+......+n(n+1)
=(1×1+1)+(2×2+2)+(3×3+3)+......(n×n+n)
=(1^2+2^2+3^2+......n^2)+(1+2+3+......n)
=n*(n+1)*(2*n+1)/6+n(n+1)/2
=n(n+1)(n+2)/3

Answer 2

令Sn=1*2+2*3+3*4+...+n(n+1)，
S1=1*2=2， 1*2*3/3=2
S2=1*2+2*3=8， 2*3*4/3=8
.......
对于n=t成立
令n=t+1，则左边=1*2+2*3+3*4+...+t(t+1)+(t+1)(t+2)=t(t+1)(t+2)/3+(t+1)(t+2)=（t+1)(t+2)(t+3)/3,
得证。