Calculate dy if y=sum k^2 * (k+1)/n(n+1)(n+2)

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

Sum k^2*(k+1) = Sum (k^3 + k^2) = Sum
k^3 + Sum k^2

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^3 = 1^3 + 2^3 + ... +
n^3

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

S3 =
[n(n+1)/2]^2

So, the numerator will
become:

Sum k^2*(k+1) = S2 +
S3

 S2 + S3 = n*(n+1)(2n+1)/6 +
[n(n+1)/2]^2

We'll multiply the second ratio by 3 and the
first by 2:

 S2 + S1 = 2n*(n+1)(2n+1)/12 +
3n^2(n+1)^2/12

We'll
factorize:

 S2 + S3 =
n(n+1)(4n+4+3n^2+3n)/12

We'll combine like
terms:

 S2 + S3 =
n(n+1)(3n^2+7n+4)/12

We'll calculate the roots of the
equation:

3n^2+7n+4 = 0

n1 =
[-7+sqrt(49-48)]/6

n1 = -1

n2
= -4/3

 S2 + S3 =  n(n+1)(n+1)(n +
4/3)/12

y = n(n+1)(n+1)(n +
4/3)/12n(n+1)(n+2)

We'll reduce like
terms:

y = (n+1)(n +
4/3)/12(n+2)

We'll remove the brackets from
numerator:

y = (n^2 + 7n/3 + 4/3)/
12(n+2)

Now, we'll calculate the first
derivative:

dy = (n^2 + 7n/3 + 4/3)'*12(n+2)-(n^2 + 7n/3 +
4/3)*[12(n+2)]'/144(n+2)^2

dy = 12(2n + 7/3)(n+2) - 12(n^2
+ 7n/3 + 4/3)/144(n+2)^2

We'll factorize by 12 and reduce
it:

dy = (2n^2 + 19n/3 + 14/3 - n^2 - 7n/3 - 4/3) /
12(n+2)^2

We'll combine like
terms:

dy = (n^2 - 4n - 10/3) /
12(n+2)^2