If `1/(2.3.4)+1/(3.4.5)+. . .+1/(100.101.102)=k/101` then `34k` is equal to

Answer

To solve the problem, we need to evaluate the series:

\[ S = \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \ldots + \frac{1}{100 \cdot 101 \cdot 102} \]

We can express each term in the series in a more manageable form. The general term can be rewritten as:

\[ \frac{1}{n(n+1)(n+2)} \]

for \( n = 2, 3, \ldots, 100 \).

### Step 1: Decompose the general term

We can use partial fraction decomposition to express:

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

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

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

Expanding this and equating coefficients will allow us to find \( A \), \( B \), and \( C \).

### Step 2: Find coefficients

Expanding the right-hand side:

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

Combining like terms:

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

Setting this equal to 1, we get 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. \( \frac{1}{2} + B + C = 0 \) → \( B + C = -\frac{1}{2} \)
2. \( \frac{3}{2} + 2B + C = 0 \) → \( 2B + C = -\frac{3}{2} \)

Now we can solve these two equations:

From \( B + C = -\frac{1}{2} \), we can express \( C \) as \( C = -\frac{1}{2} - B \).

Substituting into the second equation:

\[ 2B - \frac{1}{2} - B = -\frac{3}{2} \]

This simplifies to:

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

Substituting \( B = -1 \) back to find \( C \):

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

Thus, we have:

\[ A = \frac{1}{2}, \quad B = -1, \quad C = \frac{1}{2} \]

### Step 3: Rewrite the series

Now we can rewrite:

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

Thus, the series becomes:

\[ S = \sum_{n=2}^{100} \left( \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2} \right) \]

### Step 4: Evaluate the series

This can be simplified as:

\[ S = \frac{1}{2} \sum_{n=2}^{100} \frac{1}{n} - \sum_{n=2}^{100} \frac{1}{n+1} + \frac{1}{2} \sum_{n=2}^{100} \frac{1}{n+2} \]

The sums will telescope:

\[ S = \frac{1}{2} \left( \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{100} \right) - \left( \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{101} \right) + \frac{1}{2} \left( \frac{1}{4} + \ldots + \frac{1}{102} \right) \]

After simplification, we find:

\[ S = \frac{1}{2} \left( \frac{1}{2} - \frac{1}{101} - \frac{1}{102} \right) \]

### Step 5: Find \( k \)

We can express \( S \) in terms of \( k \):

\[ S = \frac{1}{2} \left( \frac{1}{6} - \frac{1}{101 \cdot 102} \right) \]

Setting \( S = \frac{k}{101} \) gives \( k = \frac{858}{102} \).

### Step 6: Calculate \( 34k \)

Finally, we compute:

\[ 34k = 34 \cdot \frac{858}{102} = 286 \]

Thus, the final answer is:

\[ \boxed{286} \]