Summation

Let the expression by $S$ , then we have:

$\begin{aligned} S & = \sum_{k=1}^{99} \frac{1}{k(k+1)(k+7)} \quad \quad \small \color{#3D99F6}{\text{By partial fractions (see Note)}} \\ & = \frac{1}{42} \sum_{k=1}^{99} \left( \frac{6}{k} - \frac{7}{k+1} + \frac{1}{k+7} \right) \\ & = \frac{1}{42} \sum_{k=1}^{99} \left( \frac{6}{k} - \frac{6}{k+1} - \frac{1}{k+1} + \frac{1}{k+7} \right) \\ & = \frac{1}{42} \left( \sum_{k=1}^{99} \frac{6}{k} - \sum_{k=1}^{99} \frac{6}{k+1} - \sum_{k=1}^{99} \frac{1}{k+1} + \sum_{k=1}^{99} \frac{1}{k+7} \right) \\ & = \frac{1}{42} \left( 6 \left[ \sum_{k=1}^{99} \frac{1}{k} - \sum_{k=2}^{100} \frac{1}{k} \right] - \sum_{k=2}^{100} \frac{1}{k} + \sum_{k=8}^{106} \frac{1}{k} \right) \\ & = \frac{1}{42} \left( 6 \left[ \frac{1}{1} - \frac{1}{100} \right] - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{101} + \frac{1}{102} + \frac{1}{103} + \frac{1}{104} + \frac{1}{105} + \frac{1}{106} \right) \\ & \approx \boxed{0.105} \end{aligned}$

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Partial Fractions

Assuming the following:

$\begin{aligned} \frac 1{k(k+1)(k+7)} & = \frac Ak + \frac B{k+1} + \frac C{k+7} \\ \implies 1 & = A(k+1)(k+7) + Bk(k+7) + Ck(k+1) \end{aligned}$

Putting $\begin{cases} k = 0 & \implies 7A = 1 & \implies A = \dfrac 17 \\ k = -1 & \implies -6B = 1 & \implies B = -\dfrac 16 \\ k = -7 & \implies 42C = 1 & \implies C = \dfrac 1{42} \end{cases}$

$\begin{aligned} \implies \frac 1{k(k+1)(k+7)} & = \frac 1{42} \left(\frac 6k - \frac 7{k+1} + \frac 1{k+7}\right) \end{aligned}$

I used programming in Java, this was the script.

class Pattern

{

static void pattern()

{

double sum=0,i;

for(i=1;i<=99;i++)

{

sum+=1/(i (i+1) (i+7));

}

System.out.print(sum);

}

}

I am sorry ,I don't know how to use Latex