$\frac{1}{(x+k)(x+2k)}$

First of all, we make the observation that $k\neq 0$ because if it was $k=0$ , the equation would result in a form like this,

$\frac{(n-1)}{x^2}=\frac{6}{x^2+24x+a}$

which clearly can't be an identity for any triple of $(n,k,a)$ .

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As such, we can manipulate the equation to form a telescoping sum by multiplying $1=\dfrac{k}{k}$ on L.H.S.

$\sum_{i=2}^n \left( \frac{1}{(x+(i-1)k)(x+ik)} \right)=\frac{6}{x^2+24x+a}\\ \implies \frac{1}{k}\cdot \sum_{i=2}^n \left(\frac{1}{x+(i-1)k}-\frac{1}{x+ik}\right)=\frac{6}{x^2+24x+a}\\ \implies \frac{1}{k}\cdot \frac{(n-1)k}{(x+k)(x+nk)}=\frac{6}{x^2+24x+a}\\ \implies (n-1)(x^2+24x+a)=6(x+k)(x+nk)$

For this to be an identity $\forall x$ , we must have,

$(n-1)(x^2+24x+a)\equiv 6(x+k)(x+nk)$

An immediate observation gives us $n=7$ because there is no other way to make the sides equivalent as the only multiplied constants on both sides are $(n-1)$ and $6$ respectively. Now, with the value of $n$ obtained, we can make the quadratic factors equivalent easily. We proceed as follows:

$x^2+24x+a\equiv (x+k)(x+7k)\equiv x^2+8kx+7k^2\\ \implies 24=8k~\land~a=7k^2\\ \implies k=3~\land~a=7(3^2)=63$

The triple mentioned in the problem is thus $(7,3,63)$ .

$\therefore \quad \boxed{a=63}$

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

2/(x+k)(x+3k)+1/(x+3k)(x+4k)=3/(x+k)(x+4k)

And so on... 5/(x+k)(x+6k)+1/(x+k)(x+7k)=6/(x+k)(x+7k)

Here, denominator is (x+k)(x+7k)=x^2+8kx+7k^2

Compare with the question, 8k=24

k=3

Then, a= 7k^2=7*3^2=63

I think this problem it's incorrect, because the idea is to find a telescope sum that is (1/k)(1/(x+k) - 1/(x+2k)) + (1/k)(1/(x+2k) - 1/(x+3k))+...+(1/k)(1/(x+k(n-1)) - 1/(x+nk)) = (1/k)(1/(x+k) - 1/(x+nk)) = 6/(x²+24x+a), but, after that kn - k = 6, and k = 1, so, n = 7, but 24 = k²(n+1) --> absurd!...