Calculate Sum[ k(k+3)] for k = 1 to k = n.

We'll re-write the sum:Sum k(k + 3), k is an integer
number whose values are from 1 to n.

We'll remove the
brackets:

Sum k(k+3) = Sum (k^2 +
3k)

Sum (k^2 + 3k) = Sum k^2 + Sum
3k

Sum k^2 = 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6
(1)

Sum 3k = 3*Sum k

Sum k = 1
+ 2 + 3 + .... + n

The sum of the first n natural terms
is:

Sum k = n(n+1)/2

3*Sum k =
3n(n+1)/2 (2)

Sum k(k+3) = (1) +
(2)

Sum k(k+3) = n(n+1)(2n+1)/6 +
3n(n+1)/2

We'll factorize by
n(n+1)/2:

Sum k(k+3) = [n(n+1)/2]*[(2n+1)/3 +
3]

Sum k(k+3) = [n(n+1)/2]*[(2n + 1 +
9)/3]

Sum k(k+3) = [n(n+1)/2]*[(2n +
10)/3]

Sum k(k+3) = 2*[n(n+1)/2]*[(n +
5)/3]

We'll simplify and we'll
get:

Sum k(k+3) = [n(n+1)(n +
5)/3]

So, the value of the general term of
the string is:

 an = n(n+1)(n
+ 5)/3