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

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### 2 Answers

We have to find the sum of k(k+3) with k = 1 to k = n.

Now Sum[ k(k+3)]

=> Sum [ k^2 + 3k]

=> Sum [k^2] + 3*Sum [k]

For k equal to 1 to n

=> n(n+1)(2n+1)/6 + 3*n*(n+1)/2

=> n(n+1)(2n+1)/6 + (3/2)*n*(n+1)

**Therefore the required sum is n(n+1)(2n+1)/6 + (3/2)*n*(n+1).**

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**