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

### 2 Answers | Add Yours

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**

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

We presume n increases by 1 .

y = k^2*(k+1)/n(n+1)(n+2) = K / {1/n(n+1)- 1/(n+1)(n+2)}

Therefore The sum over n of K/n(n+1)+(n+2) = {1/n(n+1) - 1/(n+)(n+2)}

Therefore ,

y = k{(1/1*2-1/2*3)+(1/2*3 -1/3*4)+(1/3*4 - 1/4*5)+....(1/n*n+1 - 1/(n+1)*(n+2)}

= K{1/2 - 1/(n+1)(n+2)

y = K {0.5 - 1/(n+1)(n+2) = 0.5K - 1/(n+1)(n+2)

y = 0.5K - {K/(n+1) - K/(n+2)}

y = 0.5K -K/(n+1) +K/n+1)

Therefore y' = (0.5K)' - K/(n+1))' +(K/(n+2))'

y' = 0 -K (-1)/(n+1)^2 + K/(n+2)^2

y' = -K{ 1/(n+1)^2 - 1/(n+2)^2}