Calculate the derivative of the function f=Sum k*(k+1) / 3x*(x+1)(x+2).

To differentiate the function, we'll have to evaluate
first the sum from numerator.

Sum k*(k+1) = Sum (k^2 + k) =
Sum k^2 + Sum k

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

It is the sum of the squares of the first n terms and
the result is:

S2 =
n*(n+1)(2n+1)/6

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

It is the sum of the first n terms of an
arithmetical progression:

S1 =
n(n+1)/2

So, the numerator will
become:

Sum k*(k+1) = S2 +
S1

 S2 + S1 = n*(n+1)(2n+1)/6 +
n(n+1)/2

We'll multiply the second ratio by
3:

 S2 + S1 = n*(n+1)(2n+1)/6 +
3n(n+1)/6

We'll factorize:

 S2
+ S1 = n(n+1)(2n+1+3)/6

 S2 + S1 =
n(n+1)(2n+4)/6

 S2 + S1 =
2n(n+1)(n+2)/6

 S2 + S1 =
n(n+1)(n+2)/3

The function will
become:

f (x) = x(x+1)(x+2)/
9x*(x+1)(x+2)

We'll eliminate like
terms:

f (x) = 1/9

Now, we'll
calculate the first derivative:

f'(x) =
(1/9)'

f'(x) =
0