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,