Find the sum of the series whose nth term is given by:

(n3 – 3n)


It is given in the question that the nth term of the series,


an = n3 – 3n


Now, we need to find the sum of this series, Sn.





Note:


I. Sum of first n natural numbers, 1 + 2 +3+…n,



II. Sum of squares of first n natural numbers, 12 + 22 + 32+….n2,



III. Sum of cubes of first n natural numbers, 13 + 23 + 33 +…..n3,



IV. Sum of a constant k, N times,



So, for the given series, we need to find,



(1)


The second term in the equation,,


forms a GP, with the common ratio, r = 3.


Sum of n terms of a GP, a, ar, ar2, ar3…arn.



Here, a= 3, r = 3;


So,


(2)


Substitute (2) in (1);



Hence, the sum of the series,


1