how to do the sum of the given series---1+2+6+15+31+56+...........n terms ??please explain.

the series can be broken in the form

a_n+1= a_n+n²

1, 1+1² , 1+1²+2² , 1+1²+2²+3², ...n terms

The sum of the series is 

1 + (1+1²) + (1+1²+2²) + (1+1²+2²+3²)+ ...n terms

Taking all the 1 together it is equal to n + (1²) + (1²+2²) + (1²+2²+3²)+ ...(n-1 terms)

Becoz the sum of n consequetive natural no is n(n+1)(2n+1)/6

The series is n + {sigma frm 1 to n-1 of (n)(n+1)(2n+1)/6 }

= n(1+1/12(n-1)(n)(n+1))

Y is n + (1²) + (1²+2²) + (1²+2²+3²)+  upto (n-1 terms) ?

Taking out all 1 from each term of the series, the series takes the form 0+(1²)+(1²+2²)+(1²+2²+3²)+...(n terms)

Simply it means (1²)+(1²+2²)+(1²+2²+3²)+...+(n-1 terms)(neglecting the 0)