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

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f = sum k*(k+1)/3x*(x+1)(x+2)

S = k(k+1= k^2 + k= Sk^2 + sK

Sk^2 = 1 + 4 + 9 + 16 + ...+n = n(n+1)(2n+1)/6

Sk = 1+ 2 + 3+ ...+ n = n(n+1)/2

Then:

S = n(n+1)(2n+1)/6 + n(n+1)/2 = [n(n+1)(2n+1) + 3n(n=1)/6

= n(n+1)((2n+1) + 3)/6 = n(n+1)(2n+4)/6= n(n+1)(n+2)/3

Then:

f = x(x+1)(x+2)/3 / 3x(x+1)(x+2)

= 1/9

f(x) = 9

==> f'(x) = 0

To calculate the derivative of the sum k(k+1)/3x(x+1(x+2)

The numerator = k(k+1) = (1/3){ (k+2)(k+1)k - (k+1)k(k-1)}.

as k(k+1) { k+2- (k-1)} = (1/3)k(k+1){3}.

Therfore sum sum k(k+1) for k =1 to n

= Sum (1/3) {k+2)(k+1)k - (k+1)k)(k-1)}.

= 1/3 { (n+2)(n+1)n - (n+1)n(n-1) + (n+1)n(n-1) - n(n-1)(n-2)+.....4*3*2 - 3.2*1}

f(n) = (1/3) {(n+2)(n+1)n - 6}/3x(x+1)(x+2)

But n being descrete, f'(n) = 0.

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**