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

9 years ago

the series can be broken in the form

an+1= an+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))

9 years ago

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)