Let the sum `sum_(n=1)^(9)1/(n(n+1)(n+2))` written in the rational form be `p/q` (where p and q are co-prime), then the value of `[(q-p)/10]` is (where [.] is the greatest integer function)

Answer

To solve the problem, we need to evaluate the sum

\[ S = \sum_{n=1}^{9} \frac{1}{n(n+1)(n+2)}. \]

### Step 1: Partial Fraction Decomposition
We start by decomposing the term \(\frac{1}{n(n+1)(n+2)}\) into partial fractions:

\[ \frac{1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}. \]

Multiplying through by the denominator \(n(n+1)(n+2)\) gives:

\[ 1 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1). \]

### Step 2: Finding Coefficients
Expanding the right-hand side:

\[ 1 = A(n^2 + 3n + 2) + B(n^2 + 2n) + C(n^2 + n). \]

Combining like terms:

\[ 1 = (A + B + C)n^2 + (3A + 2B + C)n + (2A). \]

Setting coefficients equal gives us the system of equations:

1. \(A + B + C = 0\)
2. \(3A + 2B + C = 0\)
3. \(2A = 1\)

From the third equation, we find \(A = \frac{1}{2}\). Substituting \(A\) into the first two equations:

1. \(\frac{1}{2} + B + C = 0 \implies B + C = -\frac{1}{2}\)
2. \(3 \cdot \frac{1}{2} + 2B + C = 0 \implies \frac{3}{2} + 2B + C = 0\)

Substituting \(C = -\frac{1}{2} - B\) into the second equation:

\[ \frac{3}{2} + 2B - \frac{1}{2} - B = 0 \implies B + 1 = 0 \implies B = -1. \]

Then substituting \(B\) back to find \(C\):

\[ C = -\frac{1}{2} - (-1) = \frac{1}{2}. \]

Thus, we have:

\[ \frac{1}{n(n+1)(n+2)} = \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2}. \]

### Step 3: Summing the Series
Now we can rewrite the sum \(S\):

\[ S = \sum_{n=1}^{9} \left( \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2} \right). \]

This can be separated into three sums:

\[ S = \frac{1}{2} \sum_{n=1}^{9} \frac{1}{n} - \sum_{n=1}^{9} \frac{1}{n+1} + \frac{1}{2} \sum_{n=1}^{9} \frac{1}{n+2}. \]

### Step 4: Evaluating Each Sum
Calculating each part:

1. \(\sum_{n=1}^{9} \frac{1}{n} = H_9\) (the 9th harmonic number).
2. \(\sum_{n=1}^{9} \frac{1}{n+1} = H_{10} - 1\).
3. \(\sum_{n=1}^{9} \frac{1}{n+2} = H_{11} - 1 - \frac{1}{2}\).

Thus, we have:

\[ S = \frac{1}{2} H_9 - (H_{10} - 1) + \frac{1}{2}(H_{11} - 1 - \frac{1}{2}). \]

### Step 5: Simplifying the Expression
Using the properties of harmonic numbers:

\[ H_{10} = H_9 + \frac{1}{10}, \quad H_{11} = H_{10} + \frac{1}{11} = H_9 + \frac{1}{10} + \frac{1}{11}. \]

Substituting these into the expression for \(S\) and simplifying gives:

\[ S = \frac{1}{2} H_9 - (H_9 + \frac{1}{10} - 1) + \frac{1}{2}(H_9 + \frac{1}{10} + \frac{1}{11} - 1 - \frac{1}{2}). \]

After simplification, we find:

\[ S = \frac{27}{110}. \]

### Step 6: Finding \(p\) and \(q\)
Here, \(p = 27\) and \(q = 110\). Since they are co-prime, we can calculate:

\[ q - p = 110 - 27 = 83. \]

### Step 7: Final Calculation
Finally, we need to compute:

\[ \left\lfloor \frac{q - p}{10} \right\rfloor = \left\lfloor \frac{83}{10} \right\rfloor = 8. \]

Thus, the final answer is:

\[ \boxed{8}. \]