2=1+2-1
7=(1+2)+(1+2+3)-2
16=(1+2)+(1+2+3)+(1+2+3+4)-3
30=
(1+2)+(1+2+3)+(1+2+3+4)+(1+2+3+4+5)-4
50=
(1+2)+(1+2+3)+(1+2+3+4)+(1+2+3+4+5)+(1+2+3+4+5+6)-5
则第n项为数列a(n)=(n+1)n/2前n+1项之和-n-1.即[(n+1)*(n+2)*(2n+6)/12]-n-1
2=1+2-1
7=(1+2)+(1+2+3)-2
16=(1+2)+(1+2+3)+(1+2+3+4)-3
30=
(1+2)+(1+2+3)+(1+2+3+4)+(1+2+3+4+5)-4
50=
(1+2)+(1+2+3)+(1+2+3+4)+(1+2+3+4+5)+(1+2+3+4+5+6)-5
则第n项为数列a(n)=(n+1)n/2前n+1项之和-n-1.即[(n+1)*(n+2)*(2n+6)/12]-n-1