The sum of n  terms of 1  .  2  .  3+2  .  3  .  4+....... will  be

We have, Tn=n(n+1)(n+2)=n3+3n2+2n  Sum of n terms =Sn=∑n3+3∑n2+2∑n     =14n2(n+1)2+36n(n+1)(2n+1)+22n(n+1)        =nn+12nn+12+2n+1+2    =14n(n+1)(n2+5n+6)     =14n(n+1)(n+2)(n+3)